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.
My code:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root==None:
return True
label=self.check_sym(root.left,root.right)
return label
def check_sym(self,left,right):
if left==None and right!=None:
return False
if right==None and left!=None:
return False
if left==None and right==None:
return True
if left.val!=right.val:
return False
label1=self.check_sym(left.left,right.right)
label2=self.check_sym(left.right,right.left)
if label1 and label2:
return True
else:
return False