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
3
4
5
| 1
/ \
2 2
/ \ / \
3 4 4 3
|
But the following [1,2,2,null,3,null,3] is not:
1
2
3
4
5
| 1
/ \
2 2
\ \
3 3
|
Note:
Bonus points if you could solve it both recursively and iteratively.
Solution
(Symmetric-Tree.py) download
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| # 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 rec(self,leftNode,rightNode):
if leftNode == None and rightNode == None:
return True
if leftNode == None and rightNode != None:
return False
if leftNode != None and rightNode == None:
return False
if leftNode.val != rightNode.val:
return False
return self.rec(leftNode.left, rightNode.right) and self.rec(leftNode.right, rightNode.left)
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
return self.rec(root.left, root.right)
|