在递归中使用变量迭代二叉树时出错（Python）

Question

def berry_finder(t):
    """Returns True if t contains a node with the value 'berry' and 
    False otherwise.

    >>> scrat = tree('berry')
    >>> berry_finder(scrat)
    True
    >>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')])
    >>> berry_finder(sproul)
    True
    """
    if is_leaf(t):
        if label(t)=='berry':
            return True
        else :
            return False    
    else:
        if label(t)=='berry':
            return True
        else:
            for branch in branches(t):
                branch_list=[]
                branch_list+=[berry_finder(branch)]                         
            if True in branch_list:
                return True
            else:
                return False

This is the wrong code I wrote.And the function's purpose is to find if any node's value is 'berry'.

The wrong part of the code is the recursion part(ie the code under the second else )

But I think my solution is right.I create a branch_list to store all boolean value of the branches,and if any value in branch_list is True ,then return True .

I wonder whethet it is the problem of the using of branch_list ,because the branch_list may change during the different depth of recursion?

Answer 1

The problem is that you reset branch_list in each iteration of the loop, so losing the previous boolean results, and ending up with only the last result.

You need to move that out of the loop, like so (but read on to reduce the code further -- step by step):

        branch_list=[]
        for branch in branches(t):
            branch_list+=[berry_finder(branch)]

Not a problem, but it is more natural to use append here:

        branch_list=[]
        for branch in branches(t):
            branch_list.append(berry_finder(branch))

And more pythonic is to build the list in one go through map :

        branch_list = list(map(berry_finder, branches(t)))

or comprehension:

        branch_list = [berry_finder(branch) for branch in branches(t)]

But you don't actually need to collect the booleans in a list at all. Just use any :

  return any(map(berry_finder, branches(t)))

The whole function can be reduced to just this:

def berry_finder(t):
    return label(t)=='berry' or any(map(berry_finder, branches(t)))

Answer 2

You alread get answer how to use it with list

... but you don't need list for this because you can return True directly when berry_finder(branch) gives first True , and you can return False when you exit loop.

else:
    for branch in branches(t):
        if berry_finder(branch):  # 
            return True
    
    # --- after loop ---
    
    return False

---

BTW:

Instead of

    if True in branch_list:
        return True
    else:
        return False

you can use shorter

    return (True in branch_list)   # you can even skip `()`

because ... in... gives True or False

Similar

    return (label(t) == 'berry')   # you can even skip `()`

And you would replace results

    if True in branch_list:
        return False
    else:
        return True

you can use shorter

    return not (True in branch_list)

or

    return (True not in branch_list)

but it is not the same as return (False in branch_list)