Return statement from base case of recursive function (Python)

Question

I’m trying to get a better understanding of recursion using the merge sort algorithm using Python 2.7. I have written a small snippet of code to break down a list recursively. The code seems to work fine except for the last step. For the base case, the program is supposed to return a list of size 1. However, it is returning the value “none”. Where am I going wrong?

mylist = [14,88,2,14,9,123,1,5]
def concour(thelist):
    mid = len(thelist) / 2
    LeftSide = thelist[:mid]
    print LeftSide,'length is ', len(LeftSide)
    if len(LeftSide) == 1:          #the base case here
        print LeftSide        
        return LeftSide
    else:
        concour(LeftSide)    #recursive call

print concour(mylist)



"""
[14, 88, 2, 14] length is  4
[14, 88] length is  2
[14] length is  1
[14]
None
"""

Please accept this answer below. And if possible can you include at the end of the thread "is returning None" — Brian Wiley
– Brian Wiley, Commented Sep 17, 2019 at 10:37

mstbaum · Accepted Answer · 2015-01-27 19:20:53Z

2

You're missing your return statement in the recursive call.

answered Jan 27, 2015 at 19:20

mstbaum

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