[leetcode] 101. Symmetric Tree @ python

最新推荐文章于 2025-02-03 15:37:10 发布

原创最新推荐文章于 2025-02-03 15:37:10 发布 · 256 阅读

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本文介绍了一种检查二叉树是否为中心对称的方法，通过递归算法判断树的左右子树是否互为镜像，实现时间和空间复杂度分别为O(n)和O(1)。

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:

/
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)