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原题
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.
解法
递归, base case是root为空或者没有子树时, 返回True. 然后定义isMirror函数, 检查root的左右子树是否互为镜像.
Time: O(n)
Space: O(1)
代码
# 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
if root.left is None and root.right is None:
return True
return self.isMirror(root.left, root.right)
def isMirror(self, r1, r2):
if not r1 and not r2:
return True
if not r1 or not r2:
return False
if r1.val != r2.val:
return False
return self.isMirror(r1.left, r2.right) and self.isMirror(r1.right, r2.left)