Python3实现二叉树的序列化和反序列化

最新推荐文章于 2022-06-08 17:59:27 发布

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最新推荐文章于 2022-06-08 17:59:27 发布

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分类专栏： python后端算法文章标签：二叉树二叉树序列化二叉树反序列化 python3 python

本文链接：https://blog.youkuaiyun.com/weixin_43249914/article/details/84143308

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本文介绍了一种使用Python3实现的二叉树序列化和反序列化的方法。通过将二叉树转换为字符串进行保存，再从字符串还原成二叉树，实现了数据的有效存储和读取。代码示例展示了序列化和反序列化的具体实现。

Python3实现二叉树的序列化和反序列化

python

# !/usr/bin/env python
# -*- coding: utf-8 -*-
"""
@Author: P♂boy
@License: (C) Copyright 2013-2017, Node Supply Chain Manager Corporation Limited.
@Contact: 17647361832@163.com
@Software: Pycharm
@File: Tr.py
@Time: 2018/11/16 11:36
@Desc:
"""
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None


class Solution:
    def serialize(self, root):
        # write code here
        tree, ch = [root], []
        while len(tree) > 0:
            temp = tree.pop(0)
            if temp is None:
                ch.append('# ')
            else:
                ch.append(str(temp.val) + ' ')
                tree.append(temp.left)
                tree.append(temp.right)
        return ''.join(ch)

    def deserialize(self, data):
        # write code here
        s1, i = data.split(), 0     # 用空格作为分隔符
        if s1[i] == '#':
            return None
        root = TreeNode(int(s1[i]))
        tree = [root]
        while len(tree) > 0:
            te = tree.pop(0)
            i += 1
            if s1[i] != '#':
                k = TreeNode(int(s1[i]))
                te.left = k
                tree.append(k)
            i += 1
            if s1[i] != '#':
                k = TreeNode(int(s1[i]))
                te.right = k
                tree.append(k)
        return root


T = TreeNode(8)
T.left = TreeNode(9)
T.right = TreeNode(10)
T.left.left = TreeNode(20)
T.left.right = TreeNode(20)
print(Solution().serialize(T))