leetcode 226. Invert Binary Tree 反转二叉树 + DFS深度优先搜索

Invert a binary tree.

4
/ \
2 7
/ \ / \
1 3 6 9

to

4
/ \
7 2
/ \ / \
9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

这个就是最经典的反转二叉树的实现。

直接递归即可。

代码如下：

import java.util.ArrayList;import java.util.LinkedList;import java.util.List;import java.util.Queue;/*class TreeNode {     int val;     TreeNode left;     TreeNode right;     TreeNode(int x) { val = x; }}*//* * 这就是传说中牛逼哄哄的反转二叉树的问题 * 递归解决即可 * 假如是非递归解法，要注意这可能表示满二叉树 * */public class Solution {    public TreeNode invertTree(TreeNode root)     {        if(root==null)            return root;        invert(root);        return root;            }    public void invert(TreeNode root)    {        if(root!=null)        {            TreeNode tmp=root.left;            root.left=root.right;            root.right=tmp;            invert(root.left);            invert(root.right);        }else            return;    }}

下面是C++的做法，就是一个简单的DFS深度优先遍历的简单应用

代码如下：

#include <iostream>#include <algorithm>#include <vector>#include <set>#include <map>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution{public:    TreeNode* invertTree(TreeNode* root)     {        if (root == NULL)            return root;        else        {            TreeNode* tmp = root->left;            root->left = root->right;            root->right = tmp;            invertTree(root->left);            invertTree(root->right);            return root;        }    }};