Question
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree ​`​[1,2,2,3,4,4,3]​`​ is symmetric:

`1   / \  2   2 / \ / \3  4 4  3`

But the following ​`​[1,2,2,null,3,null,3]​`​ is not:

`1   / \  2   2   \   \   3    3`

Note:
Bonus points if you could solve it both recursively and iteratively.

【复杂度】

【思路】

【代码】

`/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isSymmetric(TreeNode root) {        return helper(root,root);    }    private boolean helper(TreeNode root1,TreeNode root2){        //base case        if(root1==root2&&root1==null)            return true;        if(root1==null||root2==null)            return false;        return root1.val==root2.val&&                helper(root1.left,root2.right)&&                helper(root1.right,root2.left);    }}`