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？

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