生成所有可能的深度N的树？ [英] Generating all possible trees of depth N?

问题描述

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< = N深度树。这里有什么帮助吗？鉴于我对节点所做的每次更改都可能暴露新的子树（或删除旧的子树），因此我很难弄清楚如何递归地遍历树。

I have several different types of tree nodes, each of which may have anywhere from 0 to 5 children. I'm trying to figure out an algorithm to generate all possible trees of depth <= N. Any help here? I'm having trouble figuring out how to recursively walk the tree given that each change I make to a node may expose new subtrees (or remove old ones).

推荐答案

这是我写的一个Python程序，我认为它可以满足您的要求。给定一个起始节点，它将返回所有可能的树。从本质上讲，它可以归结为通过位操作的技巧：如果一个节点有5个子树，则有2 ⁵ = 32个可能的子树，因为每个子树可以独立地存在或不存在于子树中。

Here's a Python program I wrote up that I think does what you're asking. It'll return all of the possible trees given a starting node. Essentially, it boils down to a trick with bit manipulation: if a node has 5 children, then there are 2⁵ = 32 different possible subtrees as each child can independently be present or not present in a subtree.

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

这里是图形化的测试树的表示形式：

Here's a graphical representation of the test tree:

  1
 / \
2   3
   /|\
  4 5 6

这是运行程序的输出：

1
1 [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d就像我可以将其翻译成另一种语言一样。您没有指定，所以我使用了Python；因为我在很大程度上利用了Python的列表理解功能，所以代码在Java或C ++或其他语言上会更加冗长。

If you'd like I could translate this into a different language. You didn't specify so I used Python; the code would be a bit more verbose in Java or C++ or whatever since I took advantage of Python's list comprehensions in a big way.

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