题目描述

• 感觉和合并二叉树类似，都是很好进行递归的问题
  

代码 & 思路

• 翻转当前结点的左、右结点
• 对当前结点的左、右结点进行翻转函数
• 【自底向上】

/*** 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 {TreeNode left;public TreeNode invertTree(TreeNode root) {if(root == null){return root;}invertTree(root.left);invertTree(root.right);left = root.left;root.left = root.right;root.right = left;return root;}
}

更新版

• 自顶向下，感觉这个比较好理解

class Solution {public TreeNode invertTree(TreeNode root) {if(root == null) {return null;}TreeNode temp = root.left;root.left = root.right;root.right = temp;invertTree(root.left);invertTree(root.right);return root;}
}

• 面试前再刷一次，属实是过于清晰了

class Solution {public TreeNode mirrorTree(TreeNode root) {if(root == null) return null;TreeNode temp = root.left;root.left = root.right;root.right = temp;mirrorTree(root.left);mirrorTree(root.right);return root;}
}