//给你一个二叉树的根节点 root ， 检查它是否轴对称。 
//
// 
//
// 示例 1： 
// 
// 
//输入：root = [1,2,2,3,4,4,3]
//输出：true
// 
//
// 示例 2： 
// 
// 
//输入：root = [1,2,2,null,3,null,3]
//输出：false
// 
//
// 
//
// 提示： 
//
// 
// 树中节点数目在范围 [1, 1000] 内 
// -100 <= Node.val <= 100 
// 
//
// 
//
// 进阶：你可以运用递归和迭代两种方法解决这个问题吗？ 
//
// Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 2600 👎 0//leetcode submit region begin(Prohibit modification and deletion)import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;/*** 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) {Deque<TreeNode> deque = new LinkedList<>();deque.offerFirst(root.left);deque.offerFirst(root.right);/*仅一边为null则认为不匹配；两边非null但不匹配，则认为不匹配；两边都匹配，则继续下一次比较*/while (!deque.isEmpty()){TreeNode curLeft = deque.pollFirst();TreeNode curRight = deque.pollLast();if(curLeft == null && curRight == null){continue;}if(curLeft == null && curRight != null){return false;}if(curLeft != null && curRight == null){return false;}if(curLeft.val != curRight.val){return false;}deque.offerFirst(curLeft.right);deque.offerFirst(curLeft.left);deque.offerLast(curRight.left);deque.offerLast(curRight.right);}return true;}
}
//leetcode submit region end(Prohibit modification and deletion)