使用python将树序列化为嵌套列表

发布于 2024-12-06 02:48:33 字数 494 浏览 0 评论 0原文

我有一个像这样的二叉树类：

class BinaryTree:
def __init__(self, data, left=None, right=None):
    self.data = data
    self.left = left
    self.right = right

现在我面临着将此结构序列化为嵌套列表的任务。顺便说一句，我想到了一个从左到右的遍历函数：

def binary_tree(tree):
    if tree:
        for node_data in binary_tree(tree.left):
            yield node_data
        for node_data in binary_tree(tree.right):
            yield node_data

或者有一种通用方法将其序列化为混合嵌套结构？例如，{[]} 或 [{}]？

原文

I have a binary tree class like this:

class BinaryTree:
def __init__(self, data, left=None, right=None):
    self.data = data
    self.left = left
    self.right = right

Now I'm facing a task to serialize this structure in to a nested list. BTW, I have a left-to-right traversal function in mind:

def binary_tree(tree):
    if tree:
        for node_data in binary_tree(tree.left):
            yield node_data
        for node_data in binary_tree(tree.right):
            yield node_data

Or there is a general way to serialize it into mixed nested structure? For example, {[]}, or [{}]?

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白芷 2024-12-13 02:48:33

作为 BinaryTree 的方法：

def to_dict(self):
    data = self.data
    left = self.left
    if left is not None:
        left = left.to_dict()
    right = self.right
    if right is not None:
        right = right.to_dict()
    return {'data':data, 'left':left, 'right':right}

以及作为 BinaryTree 的类方法：

def from_dict(cls, D):
    data = D['data']
    left = D['left']
    if left is not None:
        left = cls.from_dict(left)
    right = D['right']
    if right is not None:
        right = cls.from_dict(right)
    return cls(data, left, right)

as a method of BinaryTree:

def to_dict(self):
    data = self.data
    left = self.left
    if left is not None:
        left = left.to_dict()
    right = self.right
    if right is not None:
        right = right.to_dict()
    return {'data':data, 'left':left, 'right':right}

and as a class method of BinaryTree:

def from_dict(cls, D):
    data = D['data']
    left = D['left']
    if left is not None:
        left = cls.from_dict(left)
    right = D['right']
    if right is not None:
        right = cls.from_dict(right)
    return cls(data, left, right)

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