27. 二叉树的镜像

https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-08-09 10:55:34
 * @LastEditTime: 2020-08-09 10:56:24
 * @Description: https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/
 * @FilePath: \leetcode-googtech\剑指 Offer\#27.二叉树的镜像\Solution.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 mirrorTree(TreeNode root) {
        // 判断根节点是否为空
        if(root == null) return null;
        // 初始化二叉树镜像
        TreeNode result = new TreeNode(root.val);
        // 递归交换左右子树的孩子节点
        result.left = mirrorTree(root.right);
        result.right = mirrorTree(root.left);
        // 返回结果
        return result;
    }
}

Python

'''
Author: Goog Tech
Date: 2020-08-09 10:55:41
LastEditTime: 2020-08-09 10:56:48
Description: https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/
FilePath: \leetcode-googtech\剑指 Offer\#27.二叉树的镜像\Solution.py
'''

# 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 mirrorTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        # 判断根节点是否为空
        if not root: return root
        # 交换左右子树的孩子节点
        root.left, root.right = root.right, root.left
        # 递归根节点的左右子树
        self.mirrorTree(root.left)
        self.mirrorTree(root.right)
        # 返回结果
        return root