面试题 6: 重建二叉树

一. 题目

输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树.假设输入的前序遍历和中序遍历的结果中都不含重复的数字.例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序编列序列{4,7,2,1,5,3,8,6},则重建出图所示的二叉树并输出它的头结点.

代码请到我的代码库中下载 Point2Offer

二. 代码

package week_3;
/**难度系数:****
* 剑指offer: 输入某二叉树的前序遍历和中序遍历的结果，请重建出该二节树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
* 方法:递归
* 测试用例:功能测试(树的五种情况)
* @author dingding
* Date:2017-7-1 19:00
* Declaration: All Rights Reserved!
*/
public class No6 {

public static void main(String[] args) {
test1();
System.out.println();
test2();
System.out.println();
test3();
System.out.println();
test4();
System.out.println();
test5();
System.out.println();
test6();
System.out.println();
test7();
}

//构造二叉树
private static BinaryTreeNode construct(int[] preorder,int[] inorder){
if (preorder == null || inorder == null || preorder.length!=inorder.length || inorder.length <1) {
return null;
}
return constructCore(preorder, 0,preorder.length-1,inorder,0,inorder.length-1);
}

//核心代码
private static BinaryTreeNode constructCore(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
if (preStart>preEnd) {
return null;
}
int value = preorder[preStart];
int index = inStart;
//在中序遍历的数组中找根节点的位置
while (index <=inEnd && inorder[index]!=value){
index++;
}

if (index >inEnd) {
throw new RuntimeException("Invalid input.");
}

//创建当前的根节点,并且为节点赋值
BinaryTreeNode node = new BinaryTreeNode(value);

node.left = constructCore(preorder, preStart+1, preStart+index-inStart, inorder, inStart, index-1);
node.right = constructCore(preorder, preStart+index-inStart+1, preEnd, inorder, index+1, inEnd);

return node;
}

//重建后二叉树以前序遍历的方式输出
private static void printTree(BinaryTreeNode root){
if (root!=null) {
System.out.print(root.value+" ");
printTree(root.left);
printTree(root.right);
}
}
/*====================测试用例=================*/
private static void test1() {
int[] preorder = {1, 2, 4, 7, 3, 5, 6, 8};
int[] inorder = {4, 7, 2, 1, 5, 3, 8, 6};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

private static void test2() {
int[] preorder = {1, 2, 3, 4, 5};
int[] inorder = {5, 4, 3, 2, 1};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

private static void test3() {
int[] preorder = {1, 2, 3, 4, 5};
int[] inorder = {1, 2, 3, 4, 5};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

private static void test4() {
int[] preorder = {1};
int[] inorder = {1};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

private static void test5() {
int[] preorder = {1, 2, 4, 5, 3, 6, 7};
int[] inorder = {4, 2, 5, 1, 6, 3, 7};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

private static void test6() {
construct(null, null);
}

private static void test7() {
int[] preorder = {1, 2, 4, 5, 3, 6, 7};
int[] inorder = {4, 2, 8, 1, 6, 3, 7};
BinaryTreeNode root = construct(preorder, inorder);
printTree(root);
}

}

//定义一个树
class BinaryTreeNode{
int value;
BinaryTreeNode left;
BinaryTreeNode right;

public BinaryTreeNode(int value) {
this.value = value;
this.left = null;
this.right = null;
}
}

有不妥当之处,麻烦告知:D