面试中的二叉树问题总结【Java版】
2014-10-16 13:42
776 查看
面试中的二叉树问题总结【Java版】
2014-10-16 13:423718人阅读 评论(0)
收藏
举报
本文章已收录于:
分类:
数据结构(6)
作者同类文章X
•二叉树的常见问题及其解决程序
•hash_set哈希集合容器
•list和vector有什么区别?
•经典面试题:设计包含min函数的栈,O(1)空间实现方法
更多
算法(18)
作者同类文章X
[java]
view plain
copy
print?
package Algorithms.tree;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;
/**
* REFS:
* http://blog.csdn.net/fightforyourdream/article/details/16843303 面试大总结之二:Java搞定面试中的二叉树题目
* http://blog.csdn.net/luckyxiaoqiang/article/details/7518888 轻松搞定面试中的二叉树题目
* http://www.cnblogs.com/Jax/archive/2009/12/28/1633691.html 算法大全(3) 二叉树
*
* 1. 求二叉树中的节点个数: getNodeNumRec(递归),getNodeNum(迭代)
* 2. 求二叉树的深度: getDepthRec(递归),getDepth
* 3. 前序遍历,中序遍历,后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* 4.分层遍历二叉树(按层次从上往下,从左往右): levelTraversal, levelTraversalRec(递归解法)
* 5. 将二叉查找树变为有序的双向链表: convertBST2DLLRec, convertBST2DLL
* 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel
* 7. 求二叉树中叶子节点的个数:getNodeNumLeafRec, getNodeNumLeaf
* 8. 判断两棵二叉树是否相同的树:isSameRec, isSame
* 9. 判断二叉树是不是平衡二叉树:isAVLRec
* 10. 求二叉树的镜像(破坏和不破坏原来的树两种情况):
* mirrorRec, mirrorCopyRec
* mirror, mirrorCopy
* 10.1 判断两个树是否互相镜像:isMirrorRec isMirror
* 11. 求二叉树中两个节点的最低公共祖先节点:
* LAC 求解最小公共祖先, 使用list来存储path.
* LCABstRec 递归求解BST树.
* LCARec 递归算法 .
* 12. 求二叉树中节点的最大距离:getMaxDistanceRec
* 13. 由前序遍历序列和中序遍历序列重建二叉树:rebuildBinaryTreeRec
* 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTree, isCompleteBinaryTreeRec
* 15. 找出二叉树中最长连续子串(即全部往左的连续节点,或是全部往右的连续节点)findLongest
*/
public class TreeDemo {
/*
1
/ \
2 3
/ \ \
4 5 6
*/
public static void main(String[] args) {
TreeNode r1 = new TreeNode(1);
TreeNode r2 = new TreeNode(2);
TreeNode r3 = new TreeNode(3);
TreeNode r4 = new TreeNode(4);
TreeNode r5 = new TreeNode(5);
TreeNode r6 = new TreeNode(6);
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
/*
1
/ \
2 3
/ \ \
4 5 6
*/
// TreeNode r1 = new TreeNode(10);
// TreeNode r2 = new TreeNode(6);
// TreeNode r3 = new TreeNode(14);
// TreeNode r4 = new TreeNode(4);
// TreeNode r5 = new TreeNode(8);
// TreeNode r6 = new TreeNode(16);
TreeNode r7 = new TreeNode(0);
r1.left = r2;
r1.right = r3;
r2.left = r4;
r2.right = r5;
r3.right = r6;
r4.left = r7;
TreeNode t1 = new TreeNode(10);
TreeNode t2 = new TreeNode(6);
TreeNode t3 = new TreeNode(14);
TreeNode t4 = new TreeNode(4);
24000
TreeNode t5 = new TreeNode(8);
TreeNode t6 = new TreeNode(16);
TreeNode t7 = new TreeNode(0);
TreeNode t8 = new TreeNode(0);
TreeNode t9 = new TreeNode(0);
TreeNode t10 = new TreeNode(0);
TreeNode t11 = new TreeNode(0);
t1.left = t2;
t1.right = t3;
t2.left = t4;
t2.right = t5;
t3.left = t6;
t3.right = t7;
t4.left = t8;
//t4.right = t9;
t5.right = t9;
// test distance
// t5.right = t8;
// t8.right = t9;
// t9.right = t10;
// t10.right = t11;
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
// System.out.println(LCABstRec(t1, t2, t4).val);
// System.out.println(LCABstRec(t1, t2, t6).val);
// System.out.println(LCABstRec(t1, t4, t6).val);
// System.out.println(LCABstRec(t1, t4, t7).val);
// System.out.println(LCABstRec(t1, t3, t6).val);
//
// System.out.println(LCA(t1, t2, t4).val);
// System.out.println(LCA(t1, t2, t6).val);
// System.out.println(LCA(t1, t4, t6).val);
// System.out.println(LCA(t1, t4, t7).val);
// System.out.println(LCA(t1, t3, t6).val);
// System.out.println(LCA(t1, t6, t6).val);
//System.out.println(getMaxDistanceRec(t1));
//System.out.println(isSame(r1, t1));
// System.out.println(isAVLRec(r1));
//
// preorderTraversalRec(r1);
// //mirrorRec(r1);
// //TreeNode r1Mirror = mirror(r1);
//
// TreeNode r1MirrorCopy = mirrorCopy(r1);
// System.out.println();
// //preorderTraversalRec(r1Mirror);
// preorderTraversalRec(r1MirrorCopy);
//
// System.out.println();
//
// System.out.println(isMirrorRec(r1, r1MirrorCopy));
// System.out.println(isMirror(r1, r1MirrorCopy));
//System.out.println(getNodeNumKthLevelRec(r1, 5));
//System.out.println(getNodeNumLeaf(r1));
// System.out.println(getNodeNumRec(null));
// System.out.println(getNodeNum(r1));
//System.out.println(getDepthRec(null));
// System.out.println(getDepth(r1));
//
// preorderTraversalRec(r1);
// System.out.println();
// preorderTraversal(r1);
// System.out.println();
// inorderTraversalRec(r1);
//
// System.out.println();
// inorderTraversal(r1);
// postorderTraversalRec(r1);
// System.out.println();
// postorderTraversal(r1);
// System.out.println();
// levelTraversal(r1);
//
// System.out.println();
// levelTraversalRec(r1);
// TreeNode ret = convertBST2DLLRec(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// TreeNode ret2 = convertBST2DLL(r1);
// while (ret2.right != null) {
// ret2 = ret2.right;
// }
//
// while (ret2 != null) {
// System.out.print(ret2.val + " ");
// ret2 = ret2.left;
// }
//
// TreeNode ret = convertBST2DLL(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// System.out.println();
// System.out.println(findLongest(r1));
// System.out.println();
// System.out.println(findLongest2(r1));
// test the rebuildBinaryTreeRec.
//test_rebuildBinaryTreeRec();
System.out.println(isCompleteBinaryTreeRec(t1));
System.out.println(isCompleteBinaryTree(t1));
}
public static void test_rebuildBinaryTreeRec() {
ArrayList<Integer> list1 = new ArrayList<Integer>();
list1.add(1);
list1.add(2);
list1.add(4);
list1.add(5);
list1.add(3);
list1.add(6);
list1.add(7);
list1.add(8);
ArrayList<Integer> list2 = new ArrayList<Integer>();
list2.add(4);
list2.add(2);
list2.add(5);
list2.add(1);
list2.add(3);
list2.add(7);
list2.add(6);
list2.add(8);
TreeNode root = rebuildBinaryTreeRec(list1, list2);
preorderTraversalRec(root);
System.out.println();
postorderTraversalRec(root);
}
private static class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
left = null;
right = null;
}
}
/*
* null返回0,然后把左右子树的size加上即可。
* */
public static int getNodeNumRec(TreeNode root) {
if (root == null) {
return 0;
}
return getNodeNumRec(root.left) + getNodeNumRec(root.right) + 1;
}
/**
* 求二叉树中的节点个数迭代解法O(n):基本思想同LevelOrderTraversal,
* 即用一个Queue,在Java里面可以用LinkedList来模拟
*/
public static int getNodeNum(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
int cnt = 0;
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
cnt++;
}
return cnt;
}
public static int getDepthRec(TreeNode root) {
if (root == null) {
return -1;
}
return Math.max(getDepthRec(root.left), getDepthRec(root.right)) + 1;
}
/*
* 可以用 level LevelOrderTraversal 来实现,我们用一个dummyNode来分隔不同的层,这样即可计算出实际的depth.
* 1
/ \
2 3
/ \ \
4 5 6
*
* 在队列中如此排列: 1, dummy, 2, 3, dummy, 4, 5, 5, dummy
*
*/
public static int getDepth(TreeNode root) {
if (root == null) {
return 0;
}
TreeNode dummy = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummy);
int depth = -1;
while (!q.isEmpty()) {
TreeNode curr = q.poll();
if (curr == dummy) {
depth++;
if (!q.isEmpty()) { // 使用DummyNode来区分不同的层, 如果下一层不是为空,则应该在尾部加DummyNode.
q.offer(dummy);
}
}
if (curr.left != null) {
q.offer(curr.left);
}
if (curr.right != null) {
q.offer(curr.right);
}
}
return depth;
}
/*
* 3. 前序遍历,中序遍历,后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* */
public static void preorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
System.out.print(root.val + " ");
preorderTraversalRec(root.left);
preorderTraversalRec(root.right);
}
/*
* 前序遍历,Iteration 算法. 把根节点存在stack中。
* */
public static void preorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode node = s.pop();
System.out.print(node.val + " ");
if (node.right != null) { //
s.push(node.right);
}
// 我们需要先压入右节点,再压入左节点,这样就可以先弹出左节点。
if (node.left != null) {
s.push(node.left);
}
}
}
/*
* 中序遍历
* */
public static void inorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
inorderTraversalRec(root.left);
System.out.print(root.val + " ");
inorderTraversalRec(root.right);
}
/**
* 中序遍历迭代解法 ,用栈先把根节点的所有左孩子都添加到栈内,
* 然后输出栈顶元素,再处理栈顶元素的右子树
* http://www.youtube.com/watch?v=50v1sJkjxoc *
* 还有一种方法能不用递归和栈,基于线索二叉树的方法,较麻烦以后补上
* http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/ */
public static void inorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while(true) {
// 把当前节点的左节点都push到栈中.
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// 因为此时已经没有左孩子了,所以输出栈顶元素
cur = s.pop();
System.out.print(cur.val + " ");
// 准备处理右子树
cur = cur.right;
}
}
// 后序遍历
/*
* 1
/ \
2 3
/ \ \
4 5 6
if put into the stack directly, then it should be:
1, 2, 4, 5, 3, 6 in the stack.
when pop, it should be: 6, 3, 5, 4, 2, 1
if I
* */
public static void postorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
postorderTraversalRec(root.left);
postorderTraversalRec(root.right);
System.out.print(root.val + " ");
}
/**
* 后序遍历迭代解法
* http://www.youtube.com/watch?v=hv-mJUs5mvU * http://blog.csdn.net/tang_jin2015/article/details/8545457 * 从左到右的后序 与从右到左的前序的逆序是一样的,所以就简单喽! 哈哈
* 用另外一个栈进行翻转即可喽
*/
public static void postorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> out = new Stack<TreeNode>();
s.push(root);
while(!s.isEmpty()) {
TreeNode cur = s.pop();
out.push(cur);
if (cur.left != null) {
s.push(cur.left);
}
if (cur.right != null) {
s.push(cur.right);
}
}
while(!out.isEmpty()) {
System.out.print(out.pop().val + " ");
}
}
/*
* 分层遍历二叉树(按层次从上往下,从左往右)迭代
* 其实就是广度优先搜索,使用队列实现。队列初始化,将根节点压入队列。当队列不为空,进行如下操作:弹出一个节点
* ,访问,若左子节点或右子节点不为空,将其压入队列
* */
public static void levelTraversal(TreeNode root) {
if (root == null) {
return;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
TreeNode cur = q.poll();
System.out.print(cur.val + " ");
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
}
}
public static void levelTraversalRec(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
levelTraversalVisit(root, 0, ret);
System.out.println(ret);
}
/**
* 分层遍历二叉树(递归)
* 很少有人会用递归去做level traversal
* 基本思想是用一个大的ArrayList,里面包含了每一层的ArrayList。
* 大的ArrayList的size和level有关系
*
* http://discuss.leetcode.com/questions/49/binary-tree-level-order-traversal#answer-container-2543 */
public static void levelTraversalVisit(TreeNode root, int level, ArrayList<ArrayList<Integer>> ret) {
if (root == null) {
return;
}
// 如果ArrayList的层数不够用, 则新添加一层
// when size = 3, level: 0, 1, 2
if (level >= ret.size()) {
ret.add(new ArrayList<Integer>());
}
// visit 当前节点
ret.get(level).add(root.val);
// 将左子树, 右子树添加到对应的层。
levelTraversalVisit(root.left, level + 1, ret);
levelTraversalVisit(root.right, level + 1, ret);
}
/*
* 题目要求:将二叉查找树转换成排序的双向链表,不能创建新节点,只调整指针。
查找树的结点定义如下:
既然是树,其定义本身就是递归的,自然用递归算法处理就很容易。将根结点的左子树和右子树转换为有序的双向链表,
然后根节点的left指针指向左子树结果的最后一个结点,同时左子树最后一个结点的right指针指向根节点;
根节点的right指针指向右子树结果的第一个结点,
同时右子树第一个结点的left指针指向根节点。
* */
public static TreeNode convertBST2DLLRec(TreeNode root) {
return convertBST2DLLRecHelp(root)[0];
}
/*
* ret[0] 代表左指针
* ret[1] 代表右指针
* */
public static TreeNode[] convertBST2DLLRecHelp(TreeNode root) {
TreeNode[] ret = new TreeNode[2];
ret[0] = null;
ret[1] = null;
if (root == null) {
return ret;
}
if (root.left != null) {
TreeNode left[] = convertBST2DLLRecHelp(root.left);
left[1].right = root; // 将左子树的尾节点连接到根
root.left = left[1];
ret[0] = left[0];
} else {
ret[0] = root; // 左节点返回root.
}
if (root.right != null) {
TreeNode right[] = convertBST2DLLRecHelp(root.right);
right[0].left = root; // 将右子树的头节点连接到根
root.right = right[0];
ret[1] = right[1];
} else {
ret[1] = root; // 右节点返回root.
}
return ret;
}
/**
* 将二叉查找树变为有序的双向链表 迭代解法
* 类似inOrder traversal的做法
*/
public static TreeNode convertBST2DLL(TreeNode root) {
while (root == null) {
return null;
}
TreeNode pre = null;
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
TreeNode head = null; // 链表头
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
// if stack is empty, just break;
if (s.isEmpty()) {
break;
}
cur = s.pop();
if (head == null) {
head = cur;
}
// link pre and cur.
cur.left = pre;
if (pre != null) {
pre.right = cur;
}
// 左节点已经处理完了,处理右节点
cur = cur.right;
pre = cur;
}
return root;
}
/*
* * 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevel(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
int level = 0;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
TreeNode dummy = new TreeNode(0);
int cnt = 0; // record the size of the level.
q.offer(dummy);
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node == dummy) {
level++;
if (level == k) {
return cnt;
}
cnt = 0; // reset the cnt;
if (q.isEmpty()) {
break;
}
q.offer(dummy);
continue;
}
cnt++;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
return 0;
}
/*
* * 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevelRec(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
if (k == 1) {
return 1;
}
// 将左子树及右子树在K层的节点个数相加.
return getNodeNumKthLevelRec(root.left, k - 1) + getNodeNumKthLevelRec(root.right, k - 1);
}
/*
* 7. getNodeNumLeafRec 把左子树和右子树的叶子节点加在一起即可
* */
public static int getNodeNumLeafRec(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return 1;
}
return getNodeNumLeafRec(root.left) + getNodeNumLeafRec(root.right);
}
/* 7. getNodeNumLeaf
* 随便使用一种遍历方法都可以,比如,中序遍历。
* inorderTraversal,判断是不是叶子节点。
* */
public static int getNodeNumLeaf(TreeNode root) {
if (root == null) {
return 0;
}
int cnt = 0;
// we can use inorderTraversal travesal to do it.
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// all the left child has been put into the stack, let's deal with the
// current node.
cur = s.pop();
if (cur.left == null && cur.right == null) {
cnt++;
}
cur = cur.right;
}
return cnt;
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 递归解法:
* (1)如果两棵二叉树都为空,返回真
* (2)如果两棵二叉树一棵为空,另一棵不为空,返回假
* (3)如果两棵二叉树都不为空,如果对应的左子树和右子树都同构返回真,其他返回假
* */
public static boolean isSameRec(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
// 1. the value of the root should be the same;
// 2. the left tree should be the same.
// 3. the right tree should be the same.
return r1.val == r2.val &&
isSameRec(r1.left, r2.left) && isSameRec(r1.right, r2.right);
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 迭代解法
* 我们直接用中序遍历来比较就好啦
* */
public static boolean isSame(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
TreeNode cur1 = r1;
TreeNode cur2 = r2;
while (true) {
while (cur1 != null && cur2 != null) {
s1.push(cur1);
s2.push(cur2);
cur1 = cur1.left;
cur2 = cur2.left;
}
if (cur1 != null || cur2 != null) {
return false;
}
if (s1.isEmpty() && s2.isEmpty()) {
break;
}
cur1 = s1.pop();
cur2 = s2.pop();
if (cur1.val != cur2.val) {
return false;
}
cur1 = cur1.right;
cur2 = cur2.right;
}
return true;
}
/*
*
* 9. 判断二叉树是不是平衡二叉树:isAVLRec
* 1. 左子树,右子树的高度差不能超过1
* 2. 左子树,右子树都是平衡二叉树。
*
*/
public static boolean isAVLRec(TreeNode root) {
if (root == null) {
return true;
}
// 左子树,右子树都必须是平衡二叉树。
if (!isAVLRec(root.left) || !isAVLRec(root.right)) {
return false;
}
int dif = Math.abs(getDepthRec(root.left) - getDepthRec(root.right));
if (dif > 1) {
return false;
}
return true;
}
/**
* 10. 求二叉树的镜像 递归解法:
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且交换它们
TreeNode tmp = root.right;
root.right = mirrorRec(root.left);
root.left = mirrorRec(tmp);
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法:
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* */
public static TreeNode mirror(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
// 交换当前节点的左右节点
TreeNode tmp = cur.left;
cur.left = cur.right;
cur.right = tmp;
// traversal 左节点,右节点。
if (cur.right != null) {
s.push(cur.right);
}
if (cur.left != null) {
s.push(cur.left);
}
}
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法:
*
* (2) 创建一个新的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* 前序遍历我们可以立刻把新建好的左右节点创建出来,比较方便
* */
public static TreeNode mirrorCopy(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> sCopy = new Stack<TreeNode>();
s.push(root);
TreeNode rootCopy = new TreeNode(root.val);
sCopy.push(rootCopy);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
TreeNode curCopy = sCopy.pop();
// traversal 左节点,右节点。
if (cur.right != null) {
// copy 在这里做比较好,因为我们可以容易地找到它的父节点
TreeNode leftCopy = new TreeNode(cur.right.val);
curCopy.left = leftCopy;
s.push(cur.right);
sCopy.push(curCopy.left);
}
if (cur.left != null) {
// copy 在这里做比较好,因为我们可以容易地找到它的父节点
TreeNode rightCopy = new TreeNode(cur.left.val);
curCopy.right = rightCopy;
s.push(cur.left);
sCopy.push(curCopy.right);
}
}
return rootCopy;
}
/**
* 10. 求二叉树的镜像 递归解法:
*
* (1) 不破坏原来的树,新建一个树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorCopyRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且把它们连接到新建的root节点。
TreeNode rootCopy = new TreeNode(root.val);
rootCopy.left = mirrorCopyRec(root.right);
rootCopy.right = mirrorCopyRec(root.left);
return rootCopy;
}
/*
* 10.1. 判断两个树是否互相镜像
* (1) 根必须同时为空,或是同时不为空
*
* 如果根不为空:
* (1).根的值一样
* (2).r1的左树是r2的右树的镜像
* (3).r1的右树是r2的左树的镜像
* */
public static boolean isMirrorRec(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空,则返回false.
if (r1 == null || r2 == null) {
return false;
}
// If both are not null, they should be:
// 1. have same value for root.
// 2. R1's left tree is the mirror of R2's right tree;
//&nb
2d744
sp;3. R2's right tree is the mirror of R1's left tree;
return r1.val == r2.val
&& isMirrorRec(r1.left, r2.right)
&& isMirrorRec(r1.right, r2.left);
}
/*
* 10.1. 判断两个树是否互相镜像 Iterator 做法
* (1) 根必须同时为空,或是同时不为空
*
* 如果根不为空:
* traversal 整个树,判断它们是不是镜像,每次都按照反向来traversal
* (1). 当前节点的值相等
* (2). 当前节点的左右节点要镜像,
* 无论是左节点,还是右节点,对应另外一棵树的镜像位置,可以同时为空,或是同时不为空,但是不可以一个为空,一个不为空。
* */
public static boolean isMirror(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空,则返回false.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(r1);
s2.push(r2);
while (!s1.isEmpty() && !s2.isEmpty()) {
TreeNode cur1 = s1.pop();
TreeNode cur2 = s2.pop();
// 弹出的节点的值必须相等
if (cur1.val != cur2.val) {
return false;
}
// tree1的左节点,tree2的右节点,可以同时不为空,也可以同时为空,否则返回false.
TreeNode left1 = cur1.left;
TreeNode right1 = cur1.right;
TreeNode left2 = cur2.left;
TreeNode right2 = cur2.right;
if (left1 != null && right2 != null) {
s1.push(left1);
s2.push(right2);
} else if (!(left1 == null && right2 == null)) {
return false;
}
// tree1的左节点,tree2的右节点,可以同时不为空,也可以同时为空,否则返回false.
if (right1 != null && left2 != null) {
s1.push(right1);
s2.push(left2);
} else if (!(right1 == null && left2 == null)) {
return false;
}
}
return true;
}
/*
* 11. 求二叉树中两个节点的最低公共祖先节点:
* Recursion Version:
* LACRec
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LACRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
// if no node is in the node, just recursively find it in LEFT and RIGHT tree.
TreeNode left = LACRec(root.left, node1, node2);
TreeNode right = LACRec(root.right, node1, node2);
if (left == null) { // If didn't found in the left tree, then just return it from right.
return right;
} else if (right == null) { // Or if didn't found in the right tree, then just return it from the left side.
return left;
}
// if both right and right found a node, just return the root as the Common Ancestor.
return root;
}
/*
* 11. 求BST中两个节点的最低公共祖先节点:
* Recursive version:
* LCABst
*
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LCABstRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
int min = Math.min(node1.val, node2.val);
int max = Math.max(node1.val, node2.val);
// if the values are smaller than the root value, just search them in the left tree.
if (root.val > max) {
return LCABstRec(root.left, node1, node2);
} else if (root.val < min) {
// if the values are larger than the root value, just search them in the right tree.
return LCABstRec(root.right, node1, node2);
}
// if root is in the middle, just return the root.
return root;
}
/*
* 解法1. 记录下path,并且比较之:
* LAC
* http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/ * */
public static TreeNode LCA(TreeNode root, TreeNode r1, TreeNode r2) {
// If the nodes have one in the root, just return the root.
if (root == null || r1 == null || r2 == null) {
return null;
}
ArrayList<TreeNode> list1 = new ArrayList<TreeNode>();
ArrayList<TreeNode> list2 = new ArrayList<TreeNode>();
boolean find1 = LCAPath(root, r1, list1);
boolean find2 = LCAPath(root, r2, list2);
// If didn't find any of the node, just return a null.
if (!find1 || !find2) {
return null;
}
// 注意: 使用Iterator 对于linkedlist可以提高性能。
// 所以 统一使用Iterator 来进行操作。
Iterator<TreeNode> iter1 = list1.iterator();
Iterator<TreeNode> iter2 = list2.iterator();
TreeNode last = null;
while (iter1.hasNext() && iter2.hasNext()) {
TreeNode tmp1 = iter1.next();
TreeNode tmp2 = iter2.next();
if (tmp1 != tmp2) {
return last;
}
last = tmp1;
}
// If never find any node which is different, means Node 1 and Node 2 are the same one.
// so just return the last one.
return last;
}
public static boolean LCAPath(TreeNode root, TreeNode node, ArrayList<TreeNode> path) {
// if didn't find, we should return a empty path.
if (root == null || node == null) {
return false;
}
// First add the root node.
path.add(root);
// if the node is in the left side.
if (root != node
&& !LCAPath(root.left, node, path)
&& !LCAPath(root.right, node, path)
) {
// Didn't find the node. should remove the node added before.
path.remove(root);
return false;
}
// found
return true;
}
/*
* * 12. 求二叉树中节点的最大距离:getMaxDistanceRec
*
* 首先我们来定义这个距离:
* 距离定义为:两个节点间边的数目.
* 如:
* 1
* / \
* 2 3
* \
* 4
* 这里最大距离定义为2,4的距离,为3.
* 求二叉树中节点的最大距离 即二叉树中相距最远的两个节点之间的距离。 (distance / diameter)
* 递归解法:
* 返回值设计:
* 返回1. 深度, 2. 当前树的最长距离
* (1) 计算左子树的深度,右子树深度,左子树独立的链条长度,右子树独立的链条长度
* (2) 最大长度为三者之最:
* a. 通过根节点的链,为左右深度+2
* b. 左子树独立链
* c. 右子树独立链。
*
* (3)递归初始条件:
* 当root == null, depth = -1.maxDistance = -1;
*
*/
public static int getMaxDistanceRec(TreeNode root) {
return getMaxDistanceRecHelp(root).maxDistance;
}
public static Result getMaxDistanceRecHelp(TreeNode root) {
Result ret = new Result(-1, -1);
if (root == null) {
return ret;
}
Result left = getMaxDistanceRecHelp(root.left);
Result right = getMaxDistanceRecHelp(root.right);
// 深度应加1, the depth from the subtree to the root.
ret.depth = Math.max(left.depth, right.depth) + 1;
// 左子树,右子树与根的距离都要加1,所以通过根节点的路径为两边深度+2
int crossLen = left.depth + right.depth + 2;
// 求出cross根的路径,及左右子树的独立路径,这三者路径的最大值。
ret.maxDistance = Math.max(left.maxDistance, right.maxDistance);
ret.maxDistance = Math.max(ret.maxDistance, crossLen);
return ret;
}
private static class Result {
int depth;
int maxDistance;
public Result(int depth, int maxDistance) {
this.depth = depth;
this.maxDistance = maxDistance;
}
}
/*
* 13. 由前序遍历序列和中序遍历序列重建二叉树:rebuildBinaryTreeRec
* We assume that there is no duplicate in the trees.
* For example:
* 1
* / \
* 2 3
* /\ \
* 4 5 6
* /\
* 7 8
*
* PreOrder should be: 1 2 4 5 3 6 7 8
* 根 左子树 右子树
* InOrder should be: 4 2 5 1 3 7 6 8
* 左子树 根 右子树
* */
public static TreeNode rebuildBinaryTreeRec(List<Integer> preOrder, List<Integer> inOrder) {
if (preOrder == null || inOrder == null) {
return null;
}
// If the traversal is empty, just return a NULL.
if (preOrder.size() == 0 || inOrder.size() == 0) {
return null;
}
// we can get the root from the preOrder.
// Because the first one is the root.
// So we just create the root node here.
TreeNode root = new TreeNode(preOrder.get(0));
List<Integer> preOrderLeft;
List<Integer> preOrderRight;
List<Integer> inOrderLeft;
List<Integer> inOrderRight;
// 获得在 inOrder中,根的位置
int rootInIndex = inOrder.indexOf(preOrder.get(0));
preOrderLeft = preOrder.subList(1, rootInIndex + 1);
preOrderRight = preOrder.subList(rootInIndex + 1, preOrder.size());
// 得到inOrder左边的左子树
inOrderLeft = inOrder.subList(0, rootInIndex);
inOrderRight = inOrder.subList(rootInIndex + 1, inOrder.size());
// 通过 Rec 来调用生成左右子树。
root.left = rebuildBinaryTreeRec(preOrderLeft, inOrderLeft);
root.right = rebuildBinaryTreeRec(preOrderRight, inOrderRight);
return root;
}
/*
* 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTree, isCompleteBinaryTreeRec
* 进行level traversal, 一旦遇到一个节点的左节点为空,后面的节点的子节点都必须为空。而且不应该有下一行,其实就是队列中所有的
* 元素都不应该再有子元素。
* */
public static boolean isCompleteBinaryTree(TreeNode root) {
if (root == null) {
return false;
}
TreeNode dummyNode = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummyNode);
// if this is true, no node should have any child.
boolean noChild = false;
while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur == dummyNode) {
if (!q.isEmpty()) {
q.offer(dummyNode);
}
// Dummy node不需要处理。
continue;
}
if (cur.left != null) {
// 如果标记被设置,则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.left);
} else {
// 一旦某元素没有左节点或是右节点,则之后所有的元素都不应有子元素。
// 并且该元素不可以有右节点.
noChild = true;
}
if (cur.right != null) {
// 如果标记被设置,则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.right);
} else {
// 一旦某元素没有左节点或是右节点,则之后所有的元素都不应有子元素。
noChild = true;
}
}
return true;
}
/*
* 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTreeRec
*
*
* 我们可以分解为:
* CompleteBinary Tree 的条件是:
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件:
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件:
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static boolean isCompleteBinaryTreeRec(TreeNode root) {
return isCompleteBinaryTreeRecHelp(root).isCompleteBT;
}
private static class ReturnBinaryTree {
boolean isCompleteBT;
boolean isPerfectBT;
int height;
ReturnBinaryTree(boolean isCompleteBT, boolean isPerfectBT, int height) {
this.isCompleteBT = isCompleteBT;
this.isPerfectBT = isPerfectBT;
this.height = height;
}
}
/*
* 我们可以分解为:
* CompleteBinary Tree 的条件是:
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件:
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件:
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static ReturnBinaryTree isCompleteBinaryTreeRecHelp(TreeNode root) {
ReturnBinaryTree ret = new ReturnBinaryTree(true, true, -1);
if (root == null) {
return ret;
}
ReturnBinaryTree left = isCompleteBinaryTreeRecHelp(root.left);
ReturnBinaryTree right = isCompleteBinaryTreeRecHelp(root.right);
// 树的高度为左树高度,右树高度的最大值+1
ret.height = 1 + Math.max(left.height, right.height);
// set the isPerfectBT
ret.isPerfectBT = false;
if (left.isPerfectBT && right.isPerfectBT && left.height == right.height) {
ret.isPerfectBT = true;
}
// set the isCompleteBT.
/*
* CompleteBinary Tree 的条件是:
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同(其实就是本树是perfect tree)
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
* */
ret.isCompleteBT = ret.isPerfectBT
|| (left.isCompleteBT && right.isPerfectBT && left.height == right.height + 1)
|| (left.isPerfectBT && right.isCompleteBT && left.height == right.height);
return ret;
}
/*
* 15. findLongest
* 第一种解法:
* 返回左边最长,右边最长,及左子树最长,右子树最长。
* */
public static int findLongest(TreeNode root) {
if (root == null) {
return -1;
}
TreeNode l = root;
int cntL = 0;
while (l.left != null) {
cntL++;
l = l.left;
}
TreeNode r = root;
int cntR = 0;
while (r.right != null) {
cntR++;
r = r.right;
}
int lmax = findLongest(root.left);
int rmax = findLongest(root.right);
int max = Math.max(lmax, rmax);
max = Math.max(max, cntR);
max = Math.max(max, cntL);
return max;
}
/* 1
* 2 3
* 3 4
* 6 1
* 7
* 9
* 11
* 2
* 14
* */
public static int findLongest2(TreeNode root) {
int [] maxVal = new int[1];
maxVal[0] = -1;
findLongest2Help(root, maxVal);
return maxVal[0];
}
// ret:
// 0: the left side longest,
// 1: the right side longest.
static int maxLen = -1;
static int[] findLongest2Help(TreeNode root, int[] maxVal) {
int[] ret = new int[2];
if (root == null) {
ret[0] = -1;
ret[1] = -1;
return ret;
}
ret[0] = findLongest2Help(root.left, maxVal)[0] + 1;
ret[1] = findLongest2Help(root.right, maxVal)[1] + 1;
//maxLen = Math.max(maxLen, ret[0]);
//maxLen = Math.max(maxLen, ret[1]);
maxVal[0] = Math.max(maxVal[0], ret[0]);
maxVal[0] = Math.max(maxVal[0], ret[1]);
return ret;
}
}
package Algorithms.tree; import java.util.ArrayList; import java.util.Iterator; import java.util.LinkedList; import java.util.List; import java.util.Queue; import java.util.Stack; /** * REFS: * http://blog.csdn.net/fightforyourdream/article/details/16843303 面试大总结之二:Java搞定面试中的二叉树题目 * http://blog.csdn.net/luckyxiaoqiang/article/details/7518888 轻松搞定面试中的二叉树题目 * http://www.cnblogs.com/Jax/archive/2009/12/28/1633691.html 算法大全(3) 二叉树 * * 1. 求二叉树中的节点个数: getNodeNumRec(递归),getNodeNum(迭代) * 2. 求二叉树的深度: getDepthRec(递归),getDepth * 3. 前序遍历,中序遍历,后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec * (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2) * 4.分层遍历二叉树(按层次从上往下,从左往右): levelTraversal, levelTraversalRec(递归解法) * 5. 将二叉查找树变为有序的双向链表: convertBST2DLLRec, convertBST2DLL * 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel * 7. 求二叉树中叶子节点的个数:getNodeNumLeafRec, getNodeNumLeaf * 8. 判断两棵二叉树是否相同的树:isSameRec, isSame * 9. 判断二叉树是不是平衡二叉树:isAVLRec * 10. 求二叉树的镜像(破坏和不破坏原来的树两种情况): * mirrorRec, mirrorCopyRec * mirror, mirrorCopy * 10.1 判断两个树是否互相镜像:isMirrorRec isMirror * 11. 求二叉树中两个节点的最低公共祖先节点: * LAC 求解最小公共祖先, 使用list来存储path. * LCABstRec 递归求解BST树. * LCARec 递归算法 . * 12. 求二叉树中节点的最大距离:getMaxDistanceRec * 13. 由前序遍历序列和中序遍历序列重建二叉树:rebuildBinaryTreeRec * 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTree, isCompleteBinaryTreeRec * 15. 找出二叉树中最长连续子串(即全部往左的连续节点,或是全部往右的连续节点)findLongest */ public class TreeDemo { /* 1 / \ 2 3 / \ \ 4 5 6 */ public static void main(String[] args) { TreeNode r1 = new TreeNode(1); TreeNode r2 = new TreeNode(2); TreeNode r3 = new TreeNode(3); TreeNode r4 = new TreeNode(4); TreeNode r5 = new TreeNode(5); TreeNode r6 = new TreeNode(6); /* 10 / \ 6 14 / \ \ 4 8 16 / 0 */ /* 1 / \ 2 3 / \ \ 4 5 6 */ // TreeNode r1 = new TreeNode(10); // TreeNode r2 = new TreeNode(6); // TreeNode r3 = new TreeNode(14); // TreeNode r4 = new TreeNode(4); // TreeNode r5 = new TreeNode(8); // TreeNode r6 = new TreeNode(16); TreeNode r7 = new TreeNode(0); r1.left = r2; r1.right = r3; r2.left = r4; r2.right = r5; r3.right = r6; r4.left = r7; TreeNode t1 = new TreeNode(10); TreeNode t2 = new TreeNode(6); TreeNode t3 = new TreeNode(14); TreeNode t4 = new TreeNode(4); TreeNode t5 = new TreeNode(8); TreeNode t6 = new TreeNode(16); TreeNode t7 = new TreeNode(0); TreeNode t8 = new TreeNode(0); TreeNode t9 = new TreeNode(0); TreeNode t10 = new TreeNode(0); TreeNode t11 = new TreeNode(0); t1.left = t2; t1.right = t3; t2.left = t4; t2.right = t5; t3.left = t6; t3.right = t7; t4.left = t8; //t4.right = t9; t5.right = t9; // test distance // t5.right = t8; // t8.right = t9; // t9.right = t10; // t10.right = t11; /* 10 / \ 6 14 / \ \ 4 8 16 / 0 */ // System.out.println(LCABstRec(t1, t2, t4).val); // System.out.println(LCABstRec(t1, t2, t6).val); // System.out.println(LCABstRec(t1, t4, t6).val); // System.out.println(LCABstRec(t1, t4, t7).val); // System.out.println(LCABstRec(t1, t3, t6).val); // // System.out.println(LCA(t1, t2, t4).val); // System.out.println(LCA(t1, t2, t6).val); // System.out.println(LCA(t1, t4, t6).val); // System.out.println(LCA(t1, t4, t7).val); // System.out.println(LCA(t1, t3, t6).val); // System.out.println(LCA(t1, t6, t6).val); //System.out.println(getMaxDistanceRec(t1)); //System.out.println(isSame(r1, t1)); // System.out.println(isAVLRec(r1)); // // preorderTraversalRec(r1); // //mirrorRec(r1); // //TreeNode r1Mirror = mirror(r1); // // TreeNode r1MirrorCopy = mirrorCopy(r1); // System.out.println(); // //preorderTraversalRec(r1Mirror); // preorderTraversalRec(r1MirrorCopy); // // System.out.println(); // // System.out.println(isMirrorRec(r1, r1MirrorCopy)); // System.out.println(isMirror(r1, r1MirrorCopy)); //System.out.println(getNodeNumKthLevelRec(r1, 5)); //System.out.println(getNodeNumLeaf(r1)); // System.out.println(getNodeNumRec(null)); // System.out.println(getNodeNum(r1)); //System.out.println(getDepthRec(null)); // System.out.println(getDepth(r1)); // // preorderTraversalRec(r1); // System.out.println(); // preorderTraversal(r1); // System.out.println(); // inorderTraversalRec(r1); // // System.out.println(); // inorderTraversal(r1); // postorderTraversalRec(r1); // System.out.println(); // postorderTraversal(r1); // System.out.println(); // levelTraversal(r1); // // System.out.println(); // levelTraversalRec(r1); // TreeNode ret = convertBST2DLLRec(r1); // while (ret != null) { // System.out.print(ret.val + " "); // ret = ret.right; // } // TreeNode ret2 = convertBST2DLL(r1); // while (ret2.right != null) { // ret2 = ret2.right; // } // // while (ret2 != null) { // System.out.print(ret2.val + " "); // ret2 = ret2.left; // } // // TreeNode ret = convertBST2DLL(r1); // while (ret != null) { // System.out.print(ret.val + " "); // ret = ret.right; // } // System.out.println(); // System.out.println(findLongest(r1)); // System.out.println(); // System.out.println(findLongest2(r1)); // test the rebuildBinaryTreeRec. //test_rebuildBinaryTreeRec(); System.out.println(isCompleteBinaryTreeRec(t1)); System.out.println(isCompleteBinaryTree(t1)); } public static void test_rebuildBinaryTreeRec() { ArrayList<Integer> list1 = new ArrayList<Integer>(); list1.add(1); list1.add(2); list1.add(4); list1.add(5); list1.add(3); list1.add(6); list1.add(7); list1.add(8); ArrayList<Integer> list2 = new ArrayList<Integer>(); list2.add(4); list2.add(2); list2.add(5); list2.add(1); list2.add(3); list2.add(7); list2.add(6); list2.add(8); TreeNode root = rebuildBinaryTreeRec(list1, list2); preorderTraversalRec(root); System.out.println(); postorderTraversalRec(root); } private static class TreeNode{ int val; TreeNode left; TreeNode right; public TreeNode(int val){ this.val = val; left = null; right = null; } } /* * null返回0,然后把左右子树的size加上即可。 * */ public static int getNodeNumRec(TreeNode root) { if (root == null) { return 0; } return getNodeNumRec(root.left) + getNodeNumRec(root.right) + 1; } /** * 求二叉树中的节点个数迭代解法O(n):基本思想同LevelOrderTraversal, * 即用一个Queue,在Java里面可以用LinkedList来模拟 */ public static int getNodeNum(TreeNode root) { if (root == null) { return 0; } Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); int cnt = 0; while (!q.isEmpty()) { TreeNode node = q.poll(); if (node.left != null) { q.offer(node.left); } if (node.right != null) { q.offer(node.right); } cnt++; } return cnt; } public static int getDepthRec(TreeNode root) { if (root == null) { return -1; } return Math.max(getDepthRec(root.left), getDepthRec(root.right)) + 1; } /* * 可以用 level LevelOrderTraversal 来实现,我们用一个dummyNode来分隔不同的层,这样即可计算出实际的depth. * 1 / \ 2 3 / \ \ 4 5 6 * * 在队列中如此排列: 1, dummy, 2, 3, dummy, 4, 5, 5, dummy * */ public static int getDepth(TreeNode root) { if (root == null) { return 0; } TreeNode dummy = new TreeNode(0); Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); q.offer(dummy); int depth = -1; while (!q.isEmpty()) { TreeNode curr = q.poll(); if (curr == dummy) { depth++; if (!q.isEmpty()) { // 使用DummyNode来区分不同的层, 如果下一层不是为空,则应该在尾部加DummyNode. q.offer(dummy); } } if (curr.left != null) { q.offer(curr.left); } if (curr.right != null) { q.offer(curr.right); } } return depth; } /* * 3. 前序遍历,中序遍历,后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec * (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2) * */ public static void preorderTraversalRec(TreeNode root) { if (root == null) { return; } System.out.print(root.val + " "); preorderTraversalRec(root.left); preorderTraversalRec(root.right); } /* * 前序遍历,Iteration 算法. 把根节点存在stack中。 * */ public static void preorderTraversal(TreeNode root) { if (root == null) { return; } Stack<TreeNode> s = new Stack<TreeNode>(); s.push(root); while (!s.isEmpty()) { TreeNode node = s.pop(); System.out.print(node.val + " "); if (node.right != null) { // s.push(node.right); } // 我们需要先压入右节点,再压入左节点,这样就可以先弹出左节点。 if (node.left != null) { s.push(node.left); } } } /* * 中序遍历 * */ public static void inorderTraversalRec(TreeNode root) { if (root == null) { return; } inorderTraversalRec(root.left); System.out.print(root.val + " "); inorderTraversalRec(root.right); } /** * 中序遍历迭代解法 ,用栈先把根节点的所有左孩子都添加到栈内, * 然后输出栈顶元素,再处理栈顶元素的右子树 * http://www.youtube.com/watch?v=50v1sJkjxoc * * 还有一种方法能不用递归和栈,基于线索二叉树的方法,较麻烦以后补上 * http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/ */ public static void inorderTraversal(TreeNode root) { if (root == null) { return; } Stack<TreeNode> s = new Stack<TreeNode>(); TreeNode cur = root; while(true) { // 把当前节点的左节点都push到栈中. while (cur != null) { s.push(cur); cur = cur.left; } if (s.isEmpty()) { break; } // 因为此时已经没有左孩子了,所以输出栈顶元素 cur = s.pop(); System.out.print(cur.val + " "); // 准备处理右子树 cur = cur.right; } } // 后序遍历 /* * 1 / \ 2 3 / \ \ 4 5 6 if put into the stack directly, then it should be: 1, 2, 4, 5, 3, 6 in the stack. when pop, it should be: 6, 3, 5, 4, 2, 1 if I * */ public static void postorderTraversalRec(TreeNode root) { if (root == null) { return; } postorderTraversalRec(root.left); postorderTraversalRec(root.right); System.out.print(root.val + " "); } /** * 后序遍历迭代解法 * http://www.youtube.com/watch?v=hv-mJUs5mvU * http://blog.csdn.net/tang_jin2015/article/details/8545457 * 从左到右的后序 与从右到左的前序的逆序是一样的,所以就简单喽! 哈哈 * 用另外一个栈进行翻转即可喽 */ public static void postorderTraversal(TreeNode root) { if (root == null) { return; } Stack<TreeNode> s = new Stack<TreeNode>(); Stack<TreeNode> out = new Stack<TreeNode>(); s.push(root); while(!s.isEmpty()) { TreeNode cur = s.pop(); out.push(cur); if (cur.left != null) { s.push(cur.left); } if (cur.right != null) { s.push(cur.right); } } while(!out.isEmpty()) { System.out.print(out.pop().val + " "); } } /* * 分层遍历二叉树(按层次从上往下,从左往右)迭代 * 其实就是广度优先搜索,使用队列实现。队列初始化,将根节点压入队列。当队列不为空,进行如下操作:弹出一个节点 * ,访问,若左子节点或右子节点不为空,将其压入队列 * */ public static void levelTraversal(TreeNode root) { if (root == null) { return; } Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { TreeNode cur = q.poll(); System.out.print(cur.val + " "); if (cur.left != null) { q.offer(cur.left); } if (cur.right != null) { q.offer(cur.right); } } } public static void levelTraversalRec(TreeNode root) { ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>(); levelTraversalVisit(root, 0, ret); System.out.println(ret); } /** * 分层遍历二叉树(递归) * 很少有人会用递归去做level traversal * 基本思想是用一个大的ArrayList,里面包含了每一层的ArrayList。 * 大的ArrayList的size和level有关系 * * http://discuss.leetcode.com/questions/49/binary-tree-level-order-traversal#answer-container-2543 */ public static void levelTraversalVisit(TreeNode root, int level, ArrayList<ArrayList<Integer>> ret) { if (root == null) { return; } // 如果ArrayList的层数不够用, 则新添加一层 // when size = 3, level: 0, 1, 2 if (level >= ret.size()) { ret.add(new ArrayList<Integer>()); } // visit 当前节点 ret.get(level).add(root.val); // 将左子树, 右子树添加到对应的层。 levelTraversalVisit(root.left, level + 1, ret); levelTraversalVisit(root.right, level + 1, ret); } /* * 题目要求:将二叉查找树转换成排序的双向链表,不能创建新节点,只调整指针。 查找树的结点定义如下: 既然是树,其定义本身就是递归的,自然用递归算法处理就很容易。将根结点的左子树和右子树转换为有序的双向链表, 然后根节点的left指针指向左子树结果的最后一个结点,同时左子树最后一个结点的right指针指向根节点; 根节点的right指针指向右子树结果的第一个结点, 同时右子树第一个结点的left指针指向根节点。 * */ public static TreeNode convertBST2DLLRec(TreeNode root) { return convertBST2DLLRecHelp(root)[0]; } /* * ret[0] 代表左指针 * ret[1] 代表右指针 * */ public static TreeNode[] convertBST2DLLRecHelp(TreeNode root) { TreeNode[] ret = new TreeNode[2]; ret[0] = null; ret[1] = null; if (root == null) { return ret; } if (root.left != null) { TreeNode left[] = convertBST2DLLRecHelp(root.left); left[1].right = root; // 将左子树的尾节点连接到根 root.left = left[1]; ret[0] = left[0]; } else { ret[0] = root; // 左节点返回root. } if (root.right != null) { TreeNode right[] = convertBST2DLLRecHelp(root.right); right[0].left = root; // 将右子树的头节点连接到根 root.right = right[0]; ret[1] = right[1]; } else { ret[1] = root; // 右节点返回root. } return ret; } /** * 将二叉查找树变为有序的双向链表 迭代解法 * 类似inOrder traversal的做法 */ public static TreeNode convertBST2DLL(TreeNode root) { while (root == null) { return null; } TreeNode pre = null; Stack<TreeNode> s = new Stack<TreeNode>(); TreeNode cur = root; TreeNode head = null; // 链表头 while (true) { while (cur != null) { s.push(cur); cur = cur.left; } // if stack is empty, just break; if (s.isEmpty()) { break; } cur = s.pop(); if (head == null) { head = cur; } // link pre and cur. cur.left = pre; if (pre != null) { pre.right = cur; } // 左节点已经处理完了,处理右节点 cur = cur.right; pre = cur; } return root; } /* * * 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel * */ public static int getNodeNumKthLevel(TreeNode root, int k) { if (root == null || k <= 0) { return 0; } int level = 0; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); TreeNode dummy = new TreeNode(0); int cnt = 0; // record the size of the level. q.offer(dummy); while (!q.isEmpty()) { TreeNode node = q.poll(); if (node == dummy) { level++; if (level == k) { return cnt; } cnt = 0; // reset the cnt; if (q.isEmpty()) { break; } q.offer(dummy); continue; } cnt++; if (node.left != null) { q.offer(node.left); } if (node.right != null) { q.offer(node.right); } } return 0; } /* * * 6. 求二叉树第K层的节点个数:getNodeNumKthLevelRec, getNodeNumKthLevel * */ public static int getNodeNumKthLevelRec(TreeNode root, int k) { if (root == null || k <= 0) { return 0; } if (k == 1) { return 1; } // 将左子树及右子树在K层的节点个数相加. return getNodeNumKthLevelRec(root.left, k - 1) + getNodeNumKthLevelRec(root.right, k - 1); } /* * 7. getNodeNumLeafRec 把左子树和右子树的叶子节点加在一起即可 * */ public static int getNodeNumLeafRec(TreeNode root) { if (root == null) { return 0; } if (root.left == null && root.right == null) { return 1; } return getNodeNumLeafRec(root.left) + getNodeNumLeafRec(root.right); } /* 7. getNodeNumLeaf * 随便使用一种遍历方法都可以,比如,中序遍历。 * inorderTraversal,判断是不是叶子节点。 * */ public static int getNodeNumLeaf(TreeNode root) { if (root == null) { return 0; } int cnt = 0; // we can use inorderTraversal travesal to do it. Stack<TreeNode> s = new Stack<TreeNode>(); TreeNode cur = root; while (true) { while (cur != null) { s.push(cur); cur = cur.left; } if (s.isEmpty()) { break; } // all the left child has been put into the stack, let's deal with the // current node. cur = s.pop(); if (cur.left == null && cur.right == null) { cnt++; } cur = cur.right; } return cnt; } /* * 8. 判断两棵二叉树是否相同的树。 * 递归解法: * (1)如果两棵二叉树都为空,返回真 * (2)如果两棵二叉树一棵为空,另一棵不为空,返回假 * (3)如果两棵二叉树都不为空,如果对应的左子树和右子树都同构返回真,其他返回假 * */ public static boolean isSameRec(TreeNode r1, TreeNode r2) { // both are null. if (r1 == null && r2 == null) { return true; } // one is null. if (r1 == null || r2 == null) { return false; } // 1. the value of the root should be the same; // 2. the left tree should be the same. // 3. the right tree should be the same. return r1.val == r2.val && isSameRec(r1.left, r2.left) && isSameRec(r1.right, r2.right); } /* * 8. 判断两棵二叉树是否相同的树。 * 迭代解法 * 我们直接用中序遍历来比较就好啦 * */ public static boolean isSame(TreeNode r1, TreeNode r2) { // both are null. if (r1 == null && r2 == null) { return true; } // one is null. if (r1 == null || r2 == null) { return false; } Stack<TreeNode> s1 = new Stack<TreeNode>(); Stack<TreeNode> s2 = new Stack<TreeNode>(); TreeNode cur1 = r1; TreeNode cur2 = r2; while (true) { while (cur1 != null && cur2 != null) { s1.push(cur1); s2.push(cur2); cur1 = cur1.left; cur2 = cur2.left; } if (cur1 != null || cur2 != null) { return false; } if (s1.isEmpty() && s2.isEmpty()) { break; } cur1 = s1.pop(); cur2 = s2.pop(); if (cur1.val != cur2.val) { return false; } cur1 = cur1.right; cur2 = cur2.right; } return true; } /* * * 9. 判断二叉树是不是平衡二叉树:isAVLRec * 1. 左子树,右子树的高度差不能超过1 * 2. 左子树,右子树都是平衡二叉树。 * */ public static boolean isAVLRec(TreeNode root) { if (root == null) { return true; } // 左子树,右子树都必须是平衡二叉树。 if (!isAVLRec(root.left) || !isAVLRec(root.right)) { return false; } int dif = Math.abs(getDepthRec(root.left) - getDepthRec(root.right)); if (dif > 1) { return false; } return true; } /** * 10. 求二叉树的镜像 递归解法: * * (1) 破坏原来的树 * * 1 1 * / \ * 2 -----> 2 * \ / * 3 3 * */ public static TreeNode mirrorRec(TreeNode root) { if (root == null) { return null; } // 先把左右子树分别镜像,并且交换它们 TreeNode tmp = root.right; root.right = mirrorRec(root.left); root.left = mirrorRec(tmp); return root; } /** * 10. 求二叉树的镜像 Iterator解法: * * (1) 破坏原来的树 * * 1 1 * / \ * 2 -----> 2 * \ / * 3 3 * * 应该可以使用任何一种Traversal 方法。 * 我们现在可以试看看使用最简单的前序遍历。 * */ public static TreeNode mirror(TreeNode root) { if (root == null) { return null; } Stack<TreeNode> s = new Stack<TreeNode>(); s.push(root); while (!s.isEmpty()) { TreeNode cur = s.pop(); // 交换当前节点的左右节点 TreeNode tmp = cur.left; cur.left = cur.right; cur.right = tmp; // traversal 左节点,右节点。 if (cur.right != null) { s.push(cur.right); } if (cur.left != null) { s.push(cur.left); } } return root; } /** * 10. 求二叉树的镜像 Iterator解法: * * (2) 创建一个新的树 * * 1 1 * / \ * 2 -----> 2 * \ / * 3 3 * * 应该可以使用任何一种Traversal 方法。 * 我们现在可以试看看使用最简单的前序遍历。 * 前序遍历我们可以立刻把新建好的左右节点创建出来,比较方便 * */ public static TreeNode mirrorCopy(TreeNode root) { if (root == null) { return null; } Stack<TreeNode> s = new Stack<TreeNode>(); Stack<TreeNode> sCopy = new Stack<TreeNode>(); s.push(root); TreeNode rootCopy = new TreeNode(root.val); sCopy.push(rootCopy); while (!s.isEmpty()) { TreeNode cur = s.pop(); TreeNode curCopy = sCopy.pop(); // traversal 左节点,右节点。 if (cur.right != null) { // copy 在这里做比较好,因为我们可以容易地找到它的父节点 TreeNode leftCopy = new TreeNode(cur.right.val); curCopy.left = leftCopy; s.push(cur.right); sCopy.push(curCopy.left); } if (cur.left != null) { // copy 在这里做比较好,因为我们可以容易地找到它的父节点 TreeNode rightCopy = new TreeNode(cur.left.val); curCopy.right = rightCopy; s.push(cur.left); sCopy.push(curCopy.right); } } return rootCopy; } /** * 10. 求二叉树的镜像 递归解法: * * (1) 不破坏原来的树,新建一个树 * * 1 1 * / \ * 2 -----> 2 * \ / * 3 3 * */ public static TreeNode mirrorCopyRec(TreeNode root) { if (root == null) { return null; } // 先把左右子树分别镜像,并且把它们连接到新建的root节点。 TreeNode rootCopy = new TreeNode(root.val); rootCopy.left = mirrorCopyRec(root.right); rootCopy.right = mirrorCopyRec(root.left); return rootCopy; } /* * 10.1. 判断两个树是否互相镜像 * (1) 根必须同时为空,或是同时不为空 * * 如果根不为空: * (1).根的值一样 * (2).r1的左树是r2的右树的镜像 * (3).r1的右树是r2的左树的镜像 * */ public static boolean isMirrorRec(TreeNode r1, TreeNode r2){ // 如果2个树都是空树 if (r1 == null && r2 == null) { return true; } // 如果其中一个为空,则返回false. if (r1 == null || r2 == null) { return false; } // If both are not null, they should be: // 1. have same value for root. // 2. R1's left tree is the mirror of R2's right tree; // 3. R2's right tree is the mirror of R1's left tree; return r1.val == r2.val && isMirrorRec(r1.left, r2.right) && isMirrorRec(r1.right, r2.left); } /* * 10.1. 判断两个树是否互相镜像 Iterator 做法 * (1) 根必须同时为空,或是同时不为空 * * 如果根不为空: * traversal 整个树,判断它们是不是镜像,每次都按照反向来traversal * (1). 当前节点的值相等 * (2). 当前节点的左右节点要镜像, * 无论是左节点,还是右节点,对应另外一棵树的镜像位置,可以同时为空,或是同时不为空,但是不可以一个为空,一个不为空。 * */ public static boolean isMirror(TreeNode r1, TreeNode r2){ // 如果2个树都是空树 if (r1 == null && r2 == null) { return true; } // 如果其中一个为空,则返回false. if (r1 == null || r2 == null) { return false; } Stack<TreeNode> s1 = new Stack<TreeNode>(); Stack<TreeNode> s2 = new Stack<TreeNode>(); s1.push(r1); s2.push(r2); while (!s1.isEmpty() && !s2.isEmpty()) { TreeNode cur1 = s1.pop(); TreeNode cur2 = s2.pop(); // 弹出的节点的值必须相等 if (cur1.val != cur2.val) { return false; } // tree1的左节点,tree2的右节点,可以同时不为空,也可以同时为空,否则返回false. TreeNode left1 = cur1.left; TreeNode right1 = cur1.right; TreeNode left2 = cur2.left; TreeNode right2 = cur2.right; if (left1 != null && right2 != null) { s1.push(left1); s2.push(right2); } else if (!(left1 == null && right2 == null)) { return false; } // tree1的左节点,tree2的右节点,可以同时不为空,也可以同时为空,否则返回false. if (right1 != null && left2 != null) { s1.push(right1); s2.push(left2); } else if (!(right1 == null && left2 == null)) { return false; } } return true; } /* * 11. 求二叉树中两个节点的最低公共祖先节点: * Recursion Version: * LACRec * 1. If found in the left tree, return the Ancestor. * 2. If found in the right tree, return the Ancestor. * 3. If Didn't find any of the node, return null. * 4. If found both in the left and the right tree, return the root. * */ public static TreeNode LACRec(TreeNode root, TreeNode node1, TreeNode node2) { if (root == null || node1 == null || node2 == null) { return null; } // If any of the node is the root, just return the root. if (root == node1 || root == node2) { return root; } // if no node is in the node, just recursively find it in LEFT and RIGHT tree. TreeNode left = LACRec(root.left, node1, node2); TreeNode right = LACRec(root.right, node1, node2); if (left == null) { // If didn't found in the left tree, then just return it from right. return right; } else if (right == null) { // Or if didn't found in the right tree, then just return it from the left side. return left; } // if both right and right found a node, just return the root as the Common Ancestor. return root; } /* * 11. 求BST中两个节点的最低公共祖先节点: * Recursive version: * LCABst * * 1. If found in the left tree, return the Ancestor. * 2. If found in the right tree, return the Ancestor. * 3. If Didn't find any of the node, return null. * 4. If found both in the left and the right tree, return the root. * */ public static TreeNode LCABstRec(TreeNode root, TreeNode node1, TreeNode node2) { if (root == null || node1 == null || node2 == null) { return null; } // If any of the node is the root, just return the root. if (root == node1 || root == node2) { return root; } int min = Math.min(node1.val, node2.val); int max = Math.max(node1.val, node2.val); // if the values are smaller than the root value, just search them in the left tree. if (root.val > max) { return LCABstRec(root.left, node1, node2); } else if (root.val < min) { // if the values are larger than the root value, just search them in the right tree. return LCABstRec(root.right, node1, node2); } // if root is in the middle, just return the root. return root; } /* * 解法1. 记录下path,并且比较之: * LAC * http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/ * */ public static TreeNode LCA(TreeNode root, TreeNode r1, TreeNode r2) { // If the nodes have one in the root, just return the root. if (root == null || r1 == null || r2 == null) { return null; } ArrayList<TreeNode> list1 = new ArrayList<TreeNode>(); ArrayList<TreeNode> list2 = new ArrayList<TreeNode>(); boolean find1 = LCAPath(root, r1, list1); boolean find2 = LCAPath(root, r2, list2); // If didn't find any of the node, just return a null. if (!find1 || !find2) { return null; } // 注意: 使用Iterator 对于linkedlist可以提高性能。 // 所以 统一使用Iterator 来进行操作。 Iterator<TreeNode> iter1 = list1.iterator(); Iterator<TreeNode> iter2 = list2.iterator(); TreeNode last = null; while (iter1.hasNext() && iter2.hasNext()) { TreeNode tmp1 = iter1.next(); TreeNode tmp2 = iter2.next(); if (tmp1 != tmp2) { return last; } last = tmp1; } // If never find any node which is different, means Node 1 and Node 2 are the same one. // so just return the last one. return last; } public static boolean LCAPath(TreeNode root, TreeNode node, ArrayList<TreeNode> path) { // if didn't find, we should return a empty path. if (root == null || node == null) { return false; } // First add the root node. path.add(root); // if the node is in the left side. if (root != node && !LCAPath(root.left, node, path) && !LCAPath(root.right, node, path) ) { // Didn't find the node. should remove the node added before. path.remove(root); return false; } // found return true; } /* * * 12. 求二叉树中节点的最大距离:getMaxDistanceRec * * 首先我们来定义这个距离: * 距离定义为:两个节点间边的数目. * 如: * 1 * / \ * 2 3 * \ * 4 * 这里最大距离定义为2,4的距离,为3. * 求二叉树中节点的最大距离 即二叉树中相距最远的两个节点之间的距离。 (distance / diameter) * 递归解法: * 返回值设计: * 返回1. 深度, 2. 当前树的最长距离 * (1) 计算左子树的深度,右子树深度,左子树独立的链条长度,右子树独立的链条长度 * (2) 最大长度为三者之最: * a. 通过根节点的链,为左右深度+2 * b. 左子树独立链 * c. 右子树独立链。 * * (3)递归初始条件: * 当root == null, depth = -1.maxDistance = -1; * */ public static int getMaxDistanceRec(TreeNode root) { return getMaxDistanceRecHelp(root).maxDistance; } public static Result getMaxDistanceRecHelp(TreeNode root) { Result ret = new Result(-1, -1); if (root == null) { return ret; } Result left = getMaxDistanceRecHelp(root.left); Result right = getMaxDistanceRecHelp(root.right); // 深度应加1, the depth from the subtree to the root. ret.depth = Math.max(left.depth, right.depth) + 1; // 左子树,右子树与根的距离都要加1,所以通过根节点的路径为两边深度+2 int crossLen = left.depth + right.depth + 2; // 求出cross根的路径,及左右子树的独立路径,这三者路径的最大值。 ret.maxDistance = Math.max(left.maxDistance, right.maxDistance); ret.maxDistance = Math.max(ret.maxDistance, crossLen); return ret; } private static class Result { int depth; int maxDistance; public Result(int depth, int maxDistance) { this.depth = depth; this.maxDistance = maxDistance; } } /* * 13. 由前序遍历序列和中序遍历序列重建二叉树:rebuildBinaryTreeRec * We assume that there is no duplicate in the trees. * For example: * 1 * / \ * 2 3 * /\ \ * 4 5 6 * /\ * 7 8 * * PreOrder should be: 1 2 4 5 3 6 7 8 * 根 左子树 右子树 * InOrder should be: 4 2 5 1 3 7 6 8 * 左子树 根 右子树 * */ public static TreeNode rebuildBinaryTreeRec(List<Integer> preOrder, List<Integer> inOrder) { if (preOrder == null || inOrder == null) { return null; } // If the traversal is empty, just return a NULL. if (preOrder.size() == 0 || inOrder.size() == 0) { return null; } // we can get the root from the preOrder. // Because the first one is the root. // So we just create the root node here. TreeNode root = new TreeNode(preOrder.get(0)); List<Integer> preOrderLeft; List<Integer> preOrderRight; List<Integer> inOrderLeft; List<Integer> inOrderRight; // 获得在 inOrder中,根的位置 int rootInIndex = inOrder.indexOf(preOrder.get(0)); preOrderLeft = preOrder.subList(1, rootInIndex + 1); preOrderRight = preOrder.subList(rootInIndex + 1, preOrder.size()); // 得到inOrder左边的左子树 inOrderLeft = inOrder.subList(0, rootInIndex); inOrderRight = inOrder.subList(rootInIndex + 1, inOrder.size()); // 通过 Rec 来调用生成左右子树。 root.left = rebuildBinaryTreeRec(preOrderLeft, inOrderLeft); root.right = rebuildBinaryTreeRec(preOrderRight, inOrderRight); return root; } /* * 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTree, isCompleteBinaryTreeRec * 进行level traversal, 一旦遇到一个节点的左节点为空,后面的节点的子节点都必须为空。而且不应该有下一行,其实就是队列中所有的 * 元素都不应该再有子元素。 * */ public static boolean isCompleteBinaryTree(TreeNode root) { if (root == null) { return false; } TreeNode dummyNode = new TreeNode(0); Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); q.offer(dummyNode); // if this is true, no node should have any child. boolean noChild = false; while (!q.isEmpty()) { TreeNode cur = q.poll(); if (cur == dummyNode) { if (!q.isEmpty()) { q.offer(dummyNode); } // Dummy node不需要处理。 continue; } if (cur.left != null) { // 如果标记被设置,则Queue中任何元素不应再有子元素。 if (noChild) { return false; } q.offer(cur.left); } else { // 一旦某元素没有左节点或是右节点,则之后所有的元素都不应有子元素。 // 并且该元素不可以有右节点. noChild = true; } if (cur.right != null) { // 如果标记被设置,则Queue中任何元素不应再有子元素。 if (noChild) { return false; } q.offer(cur.right); } else { // 一旦某元素没有左节点或是右节点,则之后所有的元素都不应有子元素。 noChild = true; } } return true; } /* * 14. 判断二叉树是不是完全二叉树:isCompleteBinaryTreeRec * * * 我们可以分解为: * CompleteBinary Tree 的条件是: * 1. 左右子树均为Perfect binary tree, 并且两者Height相同 * 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1 * 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同 * * Base 条件: * (1) root = null: 为perfect & complete BinaryTree, Height -1; * * 而 Perfect Binary Tree的条件: * 左右子树均为Perfect Binary Tree,并且Height 相同。 * */ public static boolean isCompleteBinaryTreeRec(TreeNode root) { return isCompleteBinaryTreeRecHelp(root).isCompleteBT; } private static class ReturnBinaryTree { boolean isCompleteBT; boolean isPerfectBT; int height; ReturnBinaryTree(boolean isCompleteBT, boolean isPerfectBT, int height) { this.isCompleteBT = isCompleteBT; this.isPerfectBT = isPerfectBT; this.height = height; } } /* * 我们可以分解为: * CompleteBinary Tree 的条件是: * 1. 左右子树均为Perfect binary tree, 并且两者Height相同 * 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1 * 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同 * * Base 条件: * (1) root = null: 为perfect & complete BinaryTree, Height -1; * * 而 Perfect Binary Tree的条件: * 左右子树均为Perfect Binary Tree,并且Height 相同。 * */ public static ReturnBinaryTree isCompleteBinaryTreeRecHelp(TreeNode root) { ReturnBinaryTree ret = new ReturnBinaryTree(true, true, -1); if (root == null) { return ret; } ReturnBinaryTree left = isCompleteBinaryTreeRecHelp(root.left); ReturnBinaryTree right = isCompleteBinaryTreeRecHelp(root.right); // 树的高度为左树高度,右树高度的最大值+1 ret.height = 1 + Math.max(left.height, right.height); // set the isPerfectBT ret.isPerfectBT = false; if (left.isPerfectBT && right.isPerfectBT && left.height == right.height) { ret.isPerfectBT = true; } // set the isCompleteBT. /* * CompleteBinary Tree 的条件是: * 1. 左右子树均为Perfect binary tree, 并且两者Height相同(其实就是本树是perfect tree) * 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree,并且两者Height差1 * 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同 * */ ret.isCompleteBT = ret.isPerfectBT || (left.isCompleteBT && right.isPerfectBT && left.height == right.height + 1) || (left.isPerfectBT && right.isCompleteBT && left.height == right.height); return ret; } /* * 15. findLongest * 第一种解法: * 返回左边最长,右边最长,及左子树最长,右子树最长。 * */ public static int findLongest(TreeNode root) { if (root == null) { return -1; } TreeNode l = root; int cntL = 0; while (l.left != null) { cntL++; l = l.left; } TreeNode r = root; int cntR = 0; while (r.right != null) { cntR++; r = r.right; } int lmax = findLongest(root.left); int rmax = findLongest(root.right); int max = Math.max(lmax, rmax); max = Math.max(max, cntR); max = Math.max(max, cntL); return max; } /* 1 * 2 3 * 3 4 * 6 1 * 7 * 9 * 11 * 2 * 14 * */ public static int findLongest2(TreeNode root) { int [] maxVal = new int[1]; maxVal[0] = -1; findLongest2Help(root, maxVal); return maxVal[0]; } // ret: // 0: the left side longest, // 1: the right side longest. static int maxLen = -1; static int[] findLongest2Help(TreeNode root, int[] maxVal) { int[] ret = new int[2]; if (root == null) { ret[0] = -1; ret[1] = -1; return ret; } ret[0] = findLongest2Help(root.left, maxVal)[0] + 1; ret[1] = findLongest2Help(root.right, maxVal)[1] + 1; //maxLen = Math.max(maxLen, ret[0]); //maxLen = Math.max(maxLen, ret[1]); maxVal[0] = Math.max(maxVal[0], ret[0]); maxVal[0] = Math.max(maxVal[0], ret[1]); return ret; } }
转 http://blog.sina.com.cn/s/blog_eb52001d0102v1si.html
顶 0
踩 0
上一篇常用算法时间空间复杂度
下一篇数据库内、外连接总结
我的同类文章
数据结构(6)算法(18)
http://blog.csdn.net
•二叉树的常见问题及其解决程序2014-09-29阅读775
•list和vector有什么区别?2014-08-07阅读445
•教你透彻了解红黑树2014-08-02阅读564
•hash_set哈希集合容器2014-08-11阅读510
•经典面试题:设计包含min函数的栈,O(1)空间实现方法2014-08-07阅读416
•找工作知识储备(3)---从头说12种排序算法:原理、图解、动画视频演示、代码以及笔试面试题目中的应用2014-07-27阅读3186
相关文章推荐
- 面试中的二叉树问题总结【Java版】
- 面试中的二叉树问题总结【Java版】
- Java集合概要及面试问题总结
- 阿里JAVA开发面试常问问题总结1
- 面试大总结:Java搞定面试中的二叉树题目
- 阿里JAVA开发面试常问问题总结4
- Java面试中关于String的问题总结
- 面试大总结之二:Java搞定面试中的二叉树题目
- 面试中所有二叉树题目总结(java版)
- [置顶] JAVA面试之二叉树问题
- java面试常见问题之Hibernate总结
- java-----面试问题总结
- 面试中的问题总结++java
- 面试中所有二叉树题目总结(java版)
- 面试总结:用Java搞定二叉树方面的面试题
- 阿里JAVA开发面试常问问题总结2
- Java面试中关于String的问题总结
- 【Java面试最近遇到的问题总结】
- 面试大总结之二:Java搞定面试中的二叉树题目
- 面试笔试问题总结(五)—数组、链表、二叉树