solution/0200-0299/0226.Invert Binary Tree/README.md · GWD/leetcode

# [226. 翻转二叉树](https://leetcode-cn.com/problems/invert-binary-tree)

[English Version](/solution/0200-0299/0226.Invert%20Binary%20Tree/README_EN.md)

## 题目描述

翻转一棵二叉树。

示例：

输入：

输出：

备注: 
这个问题是受到 <a href="https://twitter.com/mxcl" target="_blank">Max Howell </a>的 <a href="https://twitter.com/mxcl/status/608682016205344768" target="_blank">原问题</a> 启发的 ：

<blockquote>谷歌：我们90％的工程师使用您编写的软件(Homebrew)，但是您却无法在面试时在白板上写出翻转二叉树这道题，这太糟糕了。</blockquote>

## 解法

### **Python3**

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root
```

### **Java**

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}
```

### **JavaScript**

```js
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function (root) {
  if (!root) return null;
  [root.left, root.right] = [root.right, root.left];
  invertTree(root.left);
  invertTree(root.right);
  return root;
};
```

### **Go**

```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func invertTree(root *TreeNode) *TreeNode {
    if (root == nil) {
        return nil
    }
    root.Left, root.Right = root.Right, root.Left
    invertTree(root.Left)
    invertTree(root.Right)
    return root
}
```

### **...**

```

GWD/leetcode

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