tree - 生成所有可能的深度 N 的树？

Question

我有几种不同类型的树节点，每个树节点可能有 0 到 5 个子节点。我正在尝试找出一种算法来生成所有可能的深度 <= N 的树。这里有什么帮助吗？考虑到我对节点所做的每次更改都可能公开新的子树(或删除旧的子树)，我无法弄清楚如何递归遍历树。

Answer 1

这是我编写的一个 Python 程序，我认为它可以满足您的要求。它将返回给定起始节点的所有可能的树。本质上，它归结为一个位操作技巧:如果一个节点有 5 个子节点，则有 2⁵ = 32 个不同的可能子树，因为每个子节点可以独立存在或不存在于子树中。

代码:

#!/usr/bin/env python

def all_combos(choices):
    """
    Given a list of items (a,b,c,...), generates all possible combinations of
    items where one item is taken from a, one from b, one from c, and so on.

    For example, all_combos([[1, 2], ["a", "b", "c"]]) yields:

        [1, "a"]
        [1, "b"]
        [1, "c"]
        [2, "a"]
        [2, "b"]
        [2, "c"]
    """
    if not choices:
        yield []
        return

    for left_choice in choices[0]:
        for right_choices in all_combos(choices[1:]):
            yield [left_choice] + right_choices

class Node:
    def __init__(self, value, children=[]):
        self.value    = value
        self.children = children

    def all_subtrees(self, max_depth):
        yield Node(self.value)

        if max_depth > 0:
            # For each child, get all of its possible sub-trees.
            child_subtrees = [list(self.children[i].all_subtrees(max_depth - 1)) for i in range(len(self.children))]

            # Now for the n children iterate through the 2^n possibilities where
            # each child's subtree is independently present or not present. The
            # i-th child is present if the i-th bit in "bits" is a 1.
            for bits in xrange(1, 2 ** len(self.children)):
                for combos in all_combos([child_subtrees[i] for i in range(len(self.children)) if bits & (1 << i) != 0]):
                    yield Node(self.value, combos)

    def __str__(self):
        """
        Display the node's value, and then its children in brackets if it has any.
        """
        if self.children:
            return "%s %s" % (self.value, self.children)
        else:
            return str(self.value)

    def __repr__(self):
        return str(self)

tree = Node(1,
[
    Node(2),
    Node(3,
    [
        Node(4),
        Node(5),
        Node(6)
    ])
])

for subtree in tree.all_subtrees(2):
    print subtree

这是测试树的图形表示:

  1 / \2   3   /|\  4 5 6

这是运行程序的输出:

11 [2]1 [3]1 [3 [4]]1 [3 [5]]1 [3 [4, 5]]1 [3 [6]]1 [3 [4, 6]]1 [3 [5, 6]]1 [3 [4, 5, 6]]1 [2, 3]1 [2, 3 [4]]1 [2, 3 [5]]1 [2, 3 [4, 5]]1 [2, 3 [6]]1 [2, 3 [4, 6]]1 [2, 3 [5, 6]]1 [2, 3 [4, 5, 6]]

如果您愿意，我可以将其翻译成其他语言。您没有指定，所以我使用了 Python；由于我在很大程度上利用了 Python 的列表理解，因此代码在 Java 或 C++ 或其他语言中会更加冗长。