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I have coded a solution to build all valid permutations of parentheses.

My code is below.

I have a question on my code based on a comment by my PEP8 checker. It said that there was no need to include the line return anywhere in the code (initially I included one). The solution works but I have never had a case using recursion where I didn't have to use the return line in the base-case.

How come I don't need it here?

The following expression initially had a return statement but I was told this was obsolete.

 if number_open == number_pairs and number_closed == number_pairs:
 print(output)
 build_parentheses.counter += 1
 return

Python 3.7 code:

"""Module builds all valid permutations of n parentheses
For example: n = 3:
 ((()))
 (()())
 (())()
 ()(())
 ()()()
Total = 5
"""


def build_parentheses(number_pairs: int,
 output="",
 number_open=0,
 number_closed=0)-> str:
 """The function that builds the parentheses. Output as a string:
 number_pairs: number of parentheses pairs user desired
 All other parameters are private
 """
 if number_open == number_pairs and number_closed == number_pairs:
 print(output)
 build_parentheses.counter += 1
 else:
 if number_open < number_pairs:
 output += "("
 build_parentheses(number_pairs, output, number_open + 1, number_closed)
 output = output[:-1]
 if number_closed < number_open and number_open:
 output += ")"
 build_parentheses(number_pairs, output, number_open, number_closed + 1)


if __name__ == "__main__":
 build_parentheses.counter = 0
 build_parentheses(5)
 print(f"=========nbuild_parentheses.counter solutions")

By comparison, in this post I made, I did use the return statement.

As mentioned here I'll provide a short answer to summarize what we discussed. A return statement only impacts code that short-circuits any remaining code that would have been called if omitted.

pseudo code snippets below

The return statement here skips snippet 2.

if (condition) 
 // .. snippet 1
 return;

// snippet 2

The return statement here is unnecessary.

if (condition) 
 // .. snippet 1
 return;
 else 
 // snippet 2

There is no need for a return statement here because when you reach the end of a function, there is an implicit return.

For example:

1 def exampleFunction():
2 if someCondition:
3 doThis()
4 else:
5 doTheOtherThing()
6

We could put a return statement after the call to doThis(), but this would make no difference to the execution of the code. When someCondition is True, we enter that code block and then call doThis(). Then, we go to line 6 and implicitly return from the function.

So after executing line 3, we jump to line 6 and implicitly return, so explicitly returning after line 3 makes no difference.