“TypeError: unsupported operand type(s) for +: 'int' and 'NoneType” 我该如何解决这个错误，你能解释为什么它不起作用

Question

The code is supposed to run through an array 'A' searching for the number of insances of an integer 'k' which it then returns. I have to use recursion and I've already completed the task using for loops. I'm not sure why I'm getting this error and want to understand why it's happening and find a solution. The guides I've read werent clear.

I've tried running 6 variations of the code. This isn't as elegant as the original but should cover every case of input.

def o(k,A):
    if len(A) == 1:
        if A[0] == k:
            return 1
        else:
            return
    else:
        if A[0] == k:
            return 1 + o(k,A[1:])
        else:
            return o(k,A[1:])

o(5,[1,2,5,3,6,5,3,5,5,4])

I would expect the function to return 4, as there are 4 instances of 5 in the array passed, but it outputs an error:

TypeError: unsupported operand type(s) for +: 'NoneType' and 'int'

Answer 1

def o1(k, A, count = 0):
    if len(A) ==0:
        return count

    if (A[0] == k):
        return o1(k, A[1:], count+1)
    else:
        return o1(k, A[1:], count)

o1(5,[1,2,5,3,6,5,3,5,5,4])

Out[1]:
4

o1(6,[1,2,5,3,6,5,3,5,5,4])

Out[2]:
1

PS This code in a case if you use only recursion. But of course most simplify solution for this task is:

 [1,2,5,3,6,5,3,5,5,4].count(5)

 Out[3]:
 4

About original question

def o(k,A):
    if len(A) == 1:
        if A[0] == k:
            return 1
        else:
            return 0   # this explicitly value 0 helps to fixed this error. When you reached the bottom and the last element didn't match the desired value it explicitly return 0
    else:
        if A[0] == k:
            return 1 + o(k,A[1:])
        else:
            return o(k,A[1:])

o(5,[1,2,5,3,6,5,3,5,5,4])

Out[12]:
4