力扣面试150题--对称二叉树

Day 41

题目描述

做法

原理：拆分为根节点的左右两棵子树，比较左子树的右和右子树的左，左子树的左和右子树的右

/*** 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) {return check(root.left,root.right);}public boolean check(TreeNode left,TreeNode right){if(left==null&&right==null){return true;}if(left==null||right==null){return false;}if(left.val!=right.val){return false;}if(!(check(left.left,right.right)&&check(left.right,right.left))){return false;}return true;}
}