题目描述

• 回溯，由最小子树的镜像，得到次小子树的镜像判断，一直回溯到根结点即可
  

代码 & 思路

/*** 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 boolean isSymmetric(TreeNode root) {// 从root开始，进行左右子树的镜像判断return isMirror(root,root);}boolean isMirror(TreeNode left, TreeNode right){// 同时为空，镜像回trueif(left == null && right == null){return true;}// 单个为空，非镜像回falseif(left == null || right == null){return false;}// 分别对当前值、当前两值的左右子树进行判断。return (left.val == right.val) && (isMirror(right.right,left.left)) && (isMirror(right.left,left.right));}
}