Question

Define a procedure, same_structure, that takes two inputs. It should output
True if the lists have the same structure, and False
otherwise. Two values, p and q have the same structure if:

Neither p or q is a list.

Both p and q are lists, they have the same number of elements, and each
element of p has the same structure as the corresponding element of q.

EDIT: To make the picture clear the following are the expected output

same_structure([1, 0, 1], [2, 1, 2])
    ---> True
same_structure([1, [0], 1], [2, 5, 3])
    ---> False
same_structure([1, [2, [3, [4, 5]]]], ['a', ['b', ['c', ['d', 'e']]]])
    ---> True
same_structure([1, [2, [3, [4, 5]]]], ['a', ['b', ['c', ['de']]]])
    ---> False 

I thought recursion would be best to solve this problem in python I have come up with the following code but its not working.

def is_list(p):
    return isinstance(p, list)

 def same_structure(a,b):
    if not is_list(a) and not is_list(b):
        return True
    elif is_list(a) and is_list(b):
        if len(a) == len(b):
            same_structure(a[1:],b[1:])
    else:
        return False

Answer 1

Recursion would be a good idea, but not the way you’ve suggested it. First off (and this may be only a typo), you don’t actually return anything here:

if len(a) == len(b):
    same_structure(a[1:],b[1:])

Second, you should recursively deal with each element, not each sublist. ie.:

if len(a) == len(b):
    for i in range(len(a)):
        if not same_structure(a[i], b[i]):
            return False
    return True
else:
    return False

Hope this helps.