226. 翻转二叉树

https://leetcode-cn.com/problems/invert-binary-tree/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-08-16 09:19:16
 * @LastEditTime: 2020-08-16 09:40:05
 * @Description: https://leetcode-cn.com/problems/invert-binary-tree/
 * @FilePath: \leetcode-googtech\#226. Invert Binary Tree\Solution.java
 * @WebSite: https://algorithm.show/
 */

/*
 * @lc app=leetcode.cn id=226 lang=java
 *
 * [226] 翻转二叉树
 */

// @lc code=start
/**
 * 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 left = invertTree(root.left);
        TreeNode right = invertTree(root.right);
        root.left = right;
        root.right = left;
        return root;
    }

    public TreeNode invertTree(TreeNode root) {
        // 判断当前节点是否为空
        if(root == null) return null;
        // 交换当前节点的左右孩子节点
        TreeNode tempNode = root.right;
        root.right = root.left;
        root.left = tempNode;
        // 递归交换当前节点的左右孩子节点
        invertTree(root.left);
        invertTree(root.right);
        // 函数返回时即表示当前节点及它的左右孩子都已交换完毕
        return root;
    }

}
// @lc code=end

Python

'''
Author: Goog Tech
Date: 2020-08-16 09:17:39
LastEditTime: 2020-08-16 09:40:13
Description: https://leetcode-cn.com/problems/invert-binary-tree/
FilePath: \leetcode-googtech\#226. Invert Binary Tree\Solution.py
WebSite: https://algorithm.show/
'''

#
# @lc app=leetcode.cn id=226 lang=python
#
# [226] 翻转二叉树
#

# @lc code=start
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root: return None
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
        return root

    def invertTree(self, root):
        # 判断当前节点是否为空
        if not root: return None
        # 交换当前节点的左右孩子节点
        root.left, root.right = root.right, root.left
        # 递归交换当前节点的左右孩子节点
        self.invertTree(root.left)
        self.invertTree(root.right)
        # 函数返回时即表示当前节点及它的左右孩子节点都已交换完毕
        return root
# @lc code=end