Creating all possible trees of depth N?

Question

I have several different types of tree nodes, each of which can have from 0 to 5 children. I am trying to find an algorithm to generate all possible depth trees <= N. Any help here? It’s hard for me to understand how to walk recursively in a tree, given that every change I make for a node can expose new subtrees (or delete old ones).

Answer 1

Here the Python program that I wrote, I think, does what you ask. It will return all possible trees when starting node. In essence, this boils down to tricks with manipulating bits: if a node has 5 children, then there are 2 ⁵ = 32 different possible subtrees, since each child can independently be present or not be present in the subtree.

code:

#!/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 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 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

Here is a graphical representation of the test tree:

  1
 / \
2 3
   / | \
  4 5 6

And here is the result of running the program:

1
12]
13]
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]]

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Answer 2

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Answer 3

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Answer 4

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