Symmetric Tree

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.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return self.isMirror(root.left, root.right)
    def isMirror(self, l, r):
        if l and r:
            return l.val == r.val and self.isMirror(l.left, r.right) and self.isMirror(l.right, r.left)
        else:
            return l == r