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.
Python
# 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
"""
return self.isMirror(root, root)
def isMirror(self, t1, t2):
if t1 is None and t2 is None:
return True
elif t1 is None or t2 is None:
return False
return t1.val == t2.val and self.isMirror(t1.right, t2.left) and self.isMirror(t1.left, t2.right)