题解 | #序列化二叉树#

序列化二叉树

https://www.nowcoder.com/practice/cf7e25aa97c04cc1a68c8f040e71fb84

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def __init__(self) -> None:
        # 序列化
        self.str = []
        # 反序列化
        self.start = -1

    def Serialize(self, root):
        # write code here
        if not root:
            self.str.append('#')
            return self.str
        self.str.append(root.val)
        self.Serialize(root.left)
        self.Serialize(root.right)
        return self.str

    def Deserialize(self, s):
        # write code here
        self.start += 1
        if s[self.start] == '#' or self.start > len(s) or not s:
            return
        node = TreeNode(int(s[self.start]))
        node.left = self.Deserialize(s)
        node.right = self.Deserialize(s)

        return node

全部评论

相关推荐

点赞 收藏 评论
分享
牛客网
牛客企业服务