In factorial using recursion when n becomes zero (n= 0) why is it not returning final result 1?

Question

def factorial(n):
  if n == 0 :
    return 1
  return n* factorial(n-1)

Here when n reaches 0 the result is 24. Why is the result not 1?

Answer 1

The best thing to do is to simulate the function on paper.

You called factorial(4).

Looking at the definition of the function what should the function do? Since n is equal to 4, it returns 4 * factorial(3).

What does factorial(3) return? 3 * factorial(2) so we have 4 * 3 * factorial(2). You keep going on this way until factorial(0) is called.

When you reach factorial(0) the function simply returns 1 without calling itself again and thus the recursion stops.

As you can see, if you call factorial(0) the result is 1. When calling factorial(4) and the recursion reaches 0 the function returns 1, the call stack unwinds and the multiplications are performed.

Answer 2

Here's a visualization

factorial(4): n != 0, so return 4 * factorial(3)
  factorial(3): n != 0, so return 3 * factorial(2)
    factorial(2); n != 0, so return 2 * factorial(1)
      factorial(1): n != 0, so return 1 * factorial(0)
        factorial(0): n == 0 so return 1
      factorial(1): 1 * factorial(0) = 1 * 1 = 1
    factorial(2): 2 * factorial(1) = 2 * 1 = 2
  factorial(3): 3 * factorial(2) = 3 * 2 = 6
factorial(4): 4 * factorial(3) = 4 * 6 = 24