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.

class Solution:    def isSymmetric(self, root):        """        :type root: TreeNode        :rtype: bool        """        def com(left,right):            if left is None and right is None:                return True            if left is None or right is None:                return False            if left.val!=right.val:                return False            return com(left.left,right.right) and com(left.right,right.left)        if not root:            return True        return com(root.left,root.right)