Leetcode: 101. Symmetric Tree

题目

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

解题思路

递归的方式判断是否对称。

判断根节点是否为空；
新建一个函数判断二者是否对称，先判断是否有一边为空，其次判断它们是值是否相等，是则进入递归，否则返回错。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def isSymmetric(self, root: TreeNode) -> bool:if root is None:return Trueelse:return self.isSym(root.left, root.right)def isSym(self, left, right):if left is None and right is None:return Trueif left is None or right is None:return Falseif left.val == right.val:tmp1 = self.isSym(left.left, right.right)tmp2 = self.isSym(left.right, right.left)return tmp1 and tmp2else:return False

第二中stack的方式
跟递归差不多的思想，但用stack来实现。


class Solution:def isSymmetric(self, root: TreeNode) -> bool:if root is None:return Truestack = [(root.left, root.right)]while stack:pair = stack.pop(0)left = pair[0]right = pair[1]if left is None and right is None:continueif left is None or right is None:return Falseif left.val==right.val:stack.insert(0, [left.left, right.right])stack.insert(0, [left.right, right.left])else:return Falsereturn True

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Leetcode: 101. Symmetric Tree

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