网站查询域名解析用extjs做的网站

给你一个二叉树的根节点 root ， 检查它是否轴对称。

输入：root = [1,2,2,3,4,4,3]
输出：true

思路：

可以先将左或右子树反转，然后对两者判断是否相同，检查是否对称

或者可以直接递归

对每一个根节点，取它的左右子树，判断左右子树的根节点值，同时递归的查找它的子树

/*** 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 TreeNode invertTree(TreeNode root) {TreeNode res = root;if(root == null) return null;TreeNode left = root.left;root.left = root.right;root.right = left;invertTree(root.left);invertTree(root.right);return res;}public boolean isSameTree(TreeNode p, TreeNode q) {if(p == null && q == null) return true;if(p == null || q == null) return false;if(p.val != q.val) return false;return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);}public boolean isSymmetric(TreeNode root) {if(root == null) return true;TreeNode res=invertTree(root.left);return isSameTree(res,root.right);}
}