Hanoi Tower Algorithm using recursion, while show the step count

Question

We all know the python code to solve Hanoi Tower problem looks like:

def hanoi(n, a, b, c):
    if n == 1:
        print a, '-->', c
        return
    hanoi(n-1, a, c, b)
    print a, '-->', c
    hanoi(n-1, b, a, c)

hanoi(4, 'A', 'B', 'C')

My question is where to add print('stepCount ={}'.format(stepCount)) to show the stepCount for each step.

****Update & Clarification***** while the output of the code above shows:

A > C
A > B
C > B
A > C
B > A
B > C
A > C

I am wondering if print statement can be added to the code to change the output to:

stepCount: 1
A > C
stepCount: 2
A > B
stepCount: 3
C > B
stepCount: 4
A > C
stepCount: 5
B > A
stepCount: 6
B > C
stepCount: 7
A > C

Answer 1

@emplatetypedef described a solution like the following:

def hanoi(n, a, b, c, count):
    if n > 0:
        count = hanoi(n-1, a, c, b, count)
        count += 1
        print('stepCount: {}'.format(count))
        print('{} --> {}'.format(a, c))
        count = hanoi(n-1, b, a, c, count)
    return count

hanoi(3, 'A', 'B', 'C', 0)

It can also be implement with a global variable (yes I know ugh...):

def hanoi(n, a, b, c):
    global stepCount
    if n > 0:
        hanoi(n-1, a, c, b)
        stepCount += 1
        print('stepCount: {}'.format(stepCount))
        print('{} --> {}'.format(a, c))
        hanoi(n-1, b, a, c)


stepCount = 0
hanoi(3, 'A', 'B', 'C')

OUTPUT (of both code snippets)

stepCount: 1
A --> C
stepCount: 2
A --> B
stepCount: 3
C --> B
stepCount: 4
A --> C
stepCount: 5
B --> A
stepCount: 6
B --> C
stepCount: 7
A --> C