How to use properly recursion and side effects in python

Question

In a tree structure, I'm trying to find all leafs of a branch. Here is what I wrote:

def leafs_of_branch(node,heads=[]):
    if len(node.children()) == 0:
        heads.append(str(node))
    else:    
        for des in node.children():
           leafs_of_branch(des)
    return heads


leafs_of_branch(node)

I don't know why but it feels wrong for me. It works but I want to know if there is a better way to use recursion without creating the heads parameter.

@closevoter -- While this will ultimately suffer from the issue raised in that classic question, that is not what this question is about. — mgilson, Jan 07 '13 at 15:57
...especially since the question specifically asks for ways to not use that parameter. — bgporter, Jan 07 '13 at 16:00

Thorsten Kranz · Answer 1 · 2013-01-08T06:14:59.790

This

def leafs_of_branch(node,heads=[]):

is always a bad idea. Better would be

def leafs_of_branch(node,heads=None):
    heads = heads or []

as otherwise you always use the same list for leafs_of_branch. In your specific case it might be o.k., but sooner or later you will run into problems.

I recommend:

def leafs_of_branch(node):
    leafs = []
    for des in node.children():
        leafs.extend(leafs_of_branch(des))
    if len(leafs)==0:
        leafs.append(str(node))
    return leafs

leafs_of_branch(node)

Instead of doing a if len(node.children()==0, I check for len(leafs) after descending into all (possibly zero) children. Thus I call node.children() only once.

mgilson · Accepted Answer · 2013-01-07T16:12:34.460

1

I believe this should work:

def leafs_of_branch(node):
    if len(node.children()) == 0:
       return [str(node)]
    else:
       x = []
       for des in node.children():
           x += leafs_of_branch(des)  #x.extend(leafs_of_branch(des)) would work too :-)
       return x

It's not very pretty and could probably be condensed a bit more, but I was trying to keep the form of your original code as much as possible to make it obvious what was going on.

Your original version won't actually work if you call it more than once because as you append to the heads list, that list will actually be saved between calls.

edited Jan 07 '13 at 16:12

answered Jan 07 '13 at 16:00

mgilson

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I'm afraid this won't work. What if the tree has, say, ten tiers? You do not have any recursion, you only return the node or all *direct* children of the node. – Thorsten Kranz Jan 07 '13 at 16:04
Also, in the `else` statement, you are appending all children, regardless of whether they are actually leaf nodes. – bogatron Jan 07 '13 at 16:05
@ThorstenKranz -- thanks. I just forgot to recurse in the `else` section (duh!). – mgilson Jan 07 '13 at 16:13
@ThorstenKranz -- Thanks for keeping me honest ;-) – mgilson Jan 07 '13 at 16:14

asermax · Answer 3 · 2013-01-07T16:30:10.370

As long as recursion goes, you are doing it right IMO; you are missing the heads paramater on the recursive call tho. The reason it's working anyway is for what other people said, default parameters are global and reused between calls.

If you want to avoid recursion altogheter, in this case you can use either a Queue or a Stack and a loop:

def leafs_of_branch(node):
    traverse = [node]
    leafs = []

    while traverse:
        node = traverse.pop()
        children = node.children()

        if children:
            traverse.extend(children)
        else:
            leafs.append(str(node))

    return leafs

bereal · Answer 4 · 2013-01-07T16:16:14.407

First of all, refrain from using mutable objects (lists, dicts etc) as default values, since default values are global and reused between the function calls:

def bad_func(val, dest=[]):
    dest.append(val)
    print dest

>>> bad_func(1)
[1]
>>> bad_func(2)
[1, 2]  # surprise!

So, the consequent calls will make something completely unexpected.

As for the recursion question, I'd re-write it like this:

from itertools import chain

def leafs_of_branch(node):
    children = node.children()
    if not children:  # better than len(children) == 0
        return (node, )

    all_leafs = (leafs_of_branch(child) for child in children)
    return chain(*all_leafs)

score 0 · Answer 5 · answered Jan 07 '13 at 16:14

You may also define recursively an iterator this way.

def leafs_of_branch(node):
    if len(node.children()) == 0:
        yield str(node)
    else:    
        for des in node.children():
            for leaf in leafs_of_branch(des):
                yield leaf

leafs = list(leafs_of_branch(node))

How to use properly recursion and side effects in python

5 Answers5