Recursive Python FizzBuzz solution

fizzbuzz-python.py

def fizzbuzz (start, end):
  if start % 15 == 0:
    print 'fizzbuzz'
  elif start % 3 == 0:
    print 'fizz'
  elif start % 5 == 0:
    print 'buzz'
  else:
    print start
  
  if start < end:
    fizzbuzz(start+1, end)
  
fizzbuzz(1, 100) 

Your implementation can be rewritten with the signature def fizbuzz(end, start=1) and called accordingly. Externally this updated call can then still be called the normal way, i.e. as fizzbuzz(100).

Anyhow, here is my recursive Python 3 implementation which doesn't require start:

def fizzbuzz_recursive(n):
    if n == 0:
        return
    fizzbuzz_recursive(n - 1)
    if n % 3 == 0 and n % 5 == 0:
        print('FizzBuzz')
    elif n % 3 == 0:
        print('Fizz')
    elif n % 5 == 0:
        print('Buzz')
    else:
        print(n)