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

**Follow up:** Solve it both recursively and iteratively.

```
/**
* 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) {
return isMirror(root,root);
}
public boolean isMirror(TreeNode l1, TreeNode l2) {
if(l1 == null && l2 == null) return true;
if(l1 == null || l2 == null) return false;
return l1.val == l2.val && isMirror(l1.left,l2.right) && isMirror(l1.right, l2.left);
}
}
```