How to enumerate the paths to every element in an arbitrarily deep n-ary tree

Question

I'm working on some Python code where I have a representation of data as n-ary trees where n is given by the user. The arity of the tree is definite but I'm struggling with an algorithm to enumerate the path from the root to each element.

For example, if I have a 3-ary tree

                                      .
                                     /|\
                                    / | \
                                   /  |  \
                                  /   |   \
                                 .    .    .
                                /|\  /|\  /|\
                               a b cd . hi j k
                                     /|\
                                    e f g

represented by the nested list

[[a, b, c], [d, [e, f, g], h], [i, j, k]]

I'd like to get a list of tuples like

[(a, 00), (b, 01), (c, 02), (d, 10), (e, 110), (f, 111), (g, 112), (h, 12), (i, 20), (j, 21), (k, 22)]

I did find a similar problem here Enumerating all paths in a tree but it's not quite what I need and I'm not sure how I might achieve the kind of enumeration I'm looking for.

It's mandatory to use the list, or can I use a data structure? — Alberto Bonsanto, May 07 '19 at 21:28
I would change the representation to a Tree structure, as it make maybe more sense ? By the way is your example correct according to your tree ? f and g is in two child node. One way with your structure would to move along the first dimension and construct the first part of the path if the value is not an array. If it is, then do the same. — Sami Tahri, May 07 '19 at 21:30
what's the output for `[[a, b, c], [d, [e, f, g], f], [g, h, i]]` ? Is it really the tree representation? — Lante Dellarovere, May 07 '19 at 21:53

score 1 · Accepted Answer · answered May 07 '19 at 22:28

I think there is a mismatch between actual tree and its representation.
If I didn't mess up with the picture it should be:

repr = [["a", "b", "c"], ["d", ["e", "f", "g"], "h"], ["i", "j", "k"]]

If your data are composed by lists like repr, can use a recursive function like this:

def tree(l, ind=""):
    for i, x in enumerate(l):
        if isinstance(x, list):
            yield from tree(x, ind + str(i))
        else:
            yield x, ind + str(i)

>>> print(list(tree(repr))
[('a', '00'), ('b', '01'), ('c', '02'), ('d', '10'), ('e', '110'), ('f', '111'), ('g', '112'), ('h', '12'), ('i', '20'), ('j', '21'), ('k', '22')]

score 0 · Answer 2 · answered May 07 '19 at 21:46

Here's a recursive method to bubble up the path to each leaf using yield from and recursion.

class Tree:
  def __init__(self, *children, data=None):
    self.data = data
    self.children = children


def find_all(root, path_to=()):
    if root is None:
        return
    if not root.children:
        yield (root.data, path_to)
    else:
        for i, node in enumerate(root.children):
            yield from find_all(node, path_to=(*path_to, i))

root = Tree(Tree(Tree(data='a'), Tree(data='b'), Tree(data='c')), Tree(Tree(data='d'), Tree(Tree(data='e'), Tree(data='f'), Tree(data='g')), Tree(data='f')), Tree(Tree(data='g'), Tree(data='h'), Tree(data='i')))

print(list(find_all(root)))
# [('a', (0, 0)), ('b', (0, 1)), ('c', (0, 2)), ('d', (1, 0)), ('e', (1, 1, 0)), ('f', (1, 1, 1)), ('g', (1, 1, 2)), ('f', (1, 2)), ('g', (2, 0)), ('h', (2, 1)), ('i', (2, 2))]

How to enumerate the paths to every element in an arbitrarily deep n-ary tree

2 Answers2