如何创建将二叉树转换为元组的函数？答案

【问题标题】：How do i create a function to convert a binary tree to a tuple?如何创建将二叉树转换为元组的函数？
【发布时间】：2021-12-01 00:56:43
【问题描述】：

我遇到了这个问题，我的任务是将元组转换为二叉树，然后将二叉树转换回元组并返回树和元组。我能够将元组转换为树，但我未能创建一个函数来执行相反的操作。我只是一个尝试学习数据结构的初学者。

这里的 parse_tuple 函数用于解析一个元组以创建一个可以正常工作的二叉树。

请帮我修复我的 tree_to_tuple 函数。任何解决逻辑的见解或提示都会很棒。

谢谢

#used for creating binary tree
class TreeNode:
    
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None

#used to parse over the tuple to create a bianry tree
def parse_tuple(data):
    
    if isinstance(data, tuple) and len(data) == 3:
        
        node = TreeNode(data[1])
        node.left = parse_tuple(data[0])
        node.right = parse_tuple(data[2])
        
    elif data is None:
        node = None
    
    else:
        node = TreeNode(data)
    
    return node

#doesnt work
def tree_to_tuple(node):
    if isinstance(node, TreeNode) and node.left is not None and node.right is not None:
        node_mid = node.key
        node_left = tree_to_tuple(node.left)
        node_right = tree_to_tuple(node.right)
    
    elif node.left is None:
        node_left = None
        
    else:
        node_right = None
        
    return (node_left, node_mid, node_right)

tree_tuple = ((1, 3, None), 2, ((None, 3, 4), 5, (6, 7, 8)))

tree2 = parse_tuple(tree_tuple)

tree_tuple2 = (1, 2, 3)
tree = parse_tuple(tree_tuple2)
print(tree_to_tuple(tree2))

如果我尝试使用 tree_to_tuple，这就是我得到的错误

文件“main.py”，第 45 行，在 tree_to_tuple 中返回（node_left，node_mid，node_right） UnboundLocalError：在赋值之前引用了局部变量“node_left”。

【问题讨论】：

嗯，你总是返回(node_left, node_mid, node_right)，但如果你经过elif 或else 分支，那么它们就不会被定义。在elif 分支中你只定义node_left 而忘记了node_mid 和node_right，在else 分支中你只定义了node_right 而忘记了node_mid 和node_left。

标签： python recursion data-structures binary-tree

【解决方案1】：

你很接近，但你的测试有点混乱。

这是一个补丁：

def tree_to_tuple(node):
    if isinstance(node, TreeNode):

        #  special case if the tree has no left and no right sub-tree
        if node.left is None and node.right is None:
            return node.key

        return (
            tree_to_tuple(node.left),
            node.key,
            tree_to_tuple(node.right)
        )
    raise ValueError('this is not a tree')

tree_tuple = ((1, 3, None), 2, ((None, 3, 4), 5, (6, 7, 8)))

tree2 = parse_tuple(tree_tuple)

tree_tuple2 = (1, 2, 3)
tree = parse_tuple(tree_tuple2)
print(tree_to_tuple(tree2))

输出：

((1, 3, None), 2, ((None, 3, 4), 5, (6, 7, 8)))

【讨论】：

非常感谢这个解决方案```
@ShivaKharbanda 不客气 :)

【解决方案2】：

这是我编写的对我有用的一段代码。

def tree_to_tuple(node):
    if isinstance(node, TreeNode):

        if node.left is not None and node.right is not None:
            node_mid = node.key
            node_left = tree_to_tuple(node.left)
            node_right = tree_to_tuple(node.right)

            return (node_left, node_mid, node_right)

        elif node.left is None and node.right is None:
            return node.key

        elif node.left is None and node.right is not None:
            node_mid = node.key
            node_right = tree_to_tuple(node.right)
            node_left = None

            return (node_left, node_mid, node_right)

        elif node.left is not None and node.right is None:
            node_mid = node.key
            node_right = None
            node_left = tree_to_tuple(node.left)

            return (node_left, node_mid, node_right)

    else:
        print("It's not a tree")

【讨论】：