题目链接：https://leetcode.com/problems/symmetric-tree/

1. 题目介绍（Symmetric Tree）

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

【Translate】： 给定二叉树的根，检查它是否是自身的镜像(即，围绕其中心对称)。

【测试用例】：

示例1：

示例2：

【条件约束】：

【跟踪】：

【Translate】： 你能用递归和迭代求解吗?

2. 题解

2.1 递归

原题解来自于 PratikSen07 的 Easy || 0 ms 100% (Fully Explained)(Java, C++, Python, JS, Python3).

这里的递归思想很简单，即判断了所有的错误情况。

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

2.2 非递归

原题解来自于 iaming 在 Recursive and non-recursive solutions in Java 的 Comment.

其主要的解题思想就是将root的左右子树逐一压入栈，然后弹出，对比该值是否相等，不相等就返回false，然后依次循环遍历所有子树。

/*** 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 boolean isSymmetric(TreeNode root) {if (root == null) return true;Stack stack = new Stack<>();stack.push(root.left);stack.push(root.right);while (!stack.empty()) {TreeNode n1 = stack.pop(), n2 = stack.pop();if (n1 == null && n2 == null) continue;if (n1 == null || n2 == null || n1.val != n2.val) return false;stack.push(n1.left);stack.push(n2.right);stack.push(n1.right);stack.push(n2.left);}return true;}
}