How to find out two lists with same structure in python?

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

Instead of same_structure(a[1:],b[1:]), you need to check item pair of a and b one by one

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)) and (len(a) == len(b)):
        return all(map(same_structure, a, b)) # Here
    return False