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

For example, this binary tree is symmetric:

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

But the following is not:

1 / \ 2 2 \ \ 3 3

**Note:**

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

confused what `"{1,#,2,3}"`

means? > read more on how binary tree is serialized on OJ.

Recursive:

public 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; } if((left==null&&right!=null)||(left!=null&right==null)){ return false; } return left.val==right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left); } }

Non-Recursive/Iteratively: