README.md

226. 翻转二叉树

English Version

题目描述

翻转一棵二叉树。

示例：

输入：

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

输出：

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

备注:
这个问题是受到 Max Howell 的 原问题 启发的 ：

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。

解法

DFS。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        def dfs(root):
            if root is None:
                return
            root.left, root.right = root.right, root.left
            dfs(root.left)
            dfs(root.right)

        dfs(root)
        return root

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        dfs(root);
        return root;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        dfs(root.left);
        dfs(root.right);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        dfs(root);
        return root;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        TreeNode* t = root->left;
        root->left = root->right;
        root->right = t;
        dfs(root->left);
        dfs(root->right);
    }
};

Go

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

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function (root) {
    function dfs(root) {
        if (!root) return;
        [root.left, root.right] = [root.right, root.left];
        dfs(root.left);
        dfs(root.right);
    }
    dfs(root);
    return root;
};

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