Having hard time understanding the function self-returning itself twice

[jagasrik]

I am trying to understand the program written here for Find all possible unique subsets from a set of distinct integers

https://www.w3resource.com/python-exerci...cise-4.php

here in the code i am not sure what

return  self.subsetsRecur(current, sset[1:]) +self.subsetsRecur(current + [sset[0]], sset[1:])

it does, when i see the output i tend to think that python first execute the all calls of

self.subsetsRecur(current, sset[1:])

then execute

 self.subsetsRecur(current + [sset[0]], sset[1:])

and concatenating the results, but i think i may be wrong, because i dont know where this results can be stored ? we are not using much of variables to store, when i see the output values for each calls it is bit confusing, can anyone please confirm or explain this part. Thank you.

class py_solution:
    def sub_sets(self, sset):
       
        return self.subsetsRecur([], sorted(sset))
    
    def subsetsRecur(self, current, sset):
        print('before if ',current,sset) 
        if sset:
            print(' if ',current,sset) 
            return  self.subsetsRecur(current, sset[1:]) +self.subsetsRecur(current + [sset[0]], sset[1:])
        print('After if ', current,sset) 
        return [current]
    
print(py_solution().sub_sets([4,5,6]))

The results with prints :

before if  [] [4, 5, 6]
 if  [] [4, 5, 6]
before if  [] [5, 6]
 if  [] [5, 6]
before if  [] [6]
 if  [] [6]
before if  [] []
After if  [] []
before if  [6] []
After if  [6] []
before if  [5] [6]
 if  [5] [6]
before if  [5] []
After if  [5] []
before if  [5, 6] []
After if  [5, 6] []
before if  [4] [5, 6]
 if  [4] [5, 6]
before if  [4] [6]
 if  [4] [6]
before if  [4] []
After if  [4] []
before if  [4, 6] []
After if  [4, 6] []
before if  [4, 5] [6]
 if  [4, 5] [6]
before if  [4, 5] []
After if  [4, 5] []
before if  [4, 5, 6] []
After if  [4, 5, 6] []
[[], [6], [5], [5, 6], [4], [4, 6], [4, 5], [4, 5, 6]]