LeetCode-102,103,107,111总结

102. Binary Tree Level Order Traversal

Problem:

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:

Given binary tree {3,9,20,#,#,15,7},

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[

[3],

[9,20],

[15,7]

]

Anwser：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
if(root==null) return result;
Queue<TreeNode> nodes = new LinkedList<TreeNode>();
nodes.add(root);
//isEmpty(),size()是collection接口的函数，Queue、List、Set等接口是collection的子接口
while(!nodes.isEmpty()){
LinkedList<Integer> innerList = new LinkedList<Integer>();
int size = nodes.size();
for(int i=0;i<size;i++){
TreeNode current = nodes.poll();
//*
innerList.add(current.val);
//
if(current.left!=null) nodes.add(current.left);
if(current.right!=null) nodes.add(current.right);
}
result.add(innerList);
}
return result;
}
}

103. Binary Tree Zigzag Level Order Traversal

Problem：

Given a binary tree, return the zigzag level order traversal of its

nodes’ values. (ie, from left to right, then right to left for the

next level and alternate between).

For example: Given binary tree {3,9,20,#,#,15,7},

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

[

[3],

[20,9],

[15,7]

]

Anwser：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result=new LinkedList<List<Integer>>();
if (root==null) return result;
Queue<TreeNode> q = new LinkedList<TreeNode>();
int level=0;
q.add(root);
while(!q.isEmpty()){
List<Integer> innerlist =new LinkedList<Integer>();
int size = q.size();
for(int i=0;i<size;i++){
TreeNode temp = q.poll();
//此处有差别
if(level%2==1) innerlist.add(0,temp.val);
else innerlist.add(temp.val);
//
if(temp.left!=null) q.add(temp.left);
if(temp.right!=null) q.add(temp.right);
}
level++;
result.add(innerlist);
}
return result;
}
}

107. Binary Tree Level Order Traversal II

Problem:

Given a binary tree, return the bottom-up level order traversal of its

nodes’ values. (ie, from left to right, level by level from leaf to

root).

For example: Given binary tree {3,9,20,#,#,15,7},

3
/ \
9  20
/ \
15  7

return its bottom-up level order traversal as:

[

[15,7],

[9,20],

[3]

]

Anwser:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
if(root==null) return result;
Queue<TreeNode> nodes = new LinkedList<TreeNode>();
nodes.add(root);
//isEmpty(),size()是collection接口的函数，Queue、List、Set等接口是collection的子接口
while(!nodes.isEmpty()){
LinkedList<Integer> innerList = new LinkedList<Integer>();
int size = nodes.size();
for(int i=0;i<size;i++){
TreeNode current = nodes.poll();
innerList.add(current.val);
if(current.left!=null) nodes.add(current.left);
if(current.right!=null) nodes.add(current.right);
}
//此处有不同
result.add(0,innerList);
//
}
return result;
}
}

111. Minimum Depth of Binary Tree My Submissions Question

Problem:

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Anwser:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int minDepth(TreeNode root) {
if(root==null) return 0;
//注意这个题的意思，求根节点到最近的叶子节点之间的距，如［1，2］最小深度为2而不是1
else if(root.left==null||root.right==null) return Math.max(minDepth(root.left),minDepth(root.right))+1;
else return Math.min(minDepth(root.left),minDepth(root.right))+1;
}
}

分析：

这均属于层序遍历（BFS），都利用了队列，只是在返回结果时有不同的顺序要求，由于列表自带list.add(index,value)这个函数（详解见/article/9636032.html），所以利用这个函数实现省了不少事。

当然BFS可以利用DFS实现，如102题用DFS实现代码如下：

public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
if (root == null) return ans;
traverse(ans, root, 0);
return ans;
}

private void traverse (ArrayList<ArrayList<Integer>> ans, TreeNode node, int level) {
if (node == null) return;
if (ans.size() >= level + 1)
ans.get(level).add(node.val);
else {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(node.val);
ans.add(temp);
}
traverse(ans, node.left, level + 1);
traverse(ans, node.right, level + 1);
}