递归的新手，找不到解决方案

Question

I'm supposed to get those results, the first two strings are the s1 and s2 and the third one is the result i should get. The first code is the one i wrote and the second one is the teacher's.

“Maria” e “Norma”: Mnao |
“Mai” e “Nor”: Mnaoir   |
“Maria” e “Noa”:False   |
“Mar” e “Noa”: MNaora   |
“Mar” e “Noar”: False

def strin(s1,s2):
    count_n(s1,s2)
    if len(s1)==len(s2) or s1[n]==s2[n]:
        if s1[0]==s2[0]:
            return ""
        else:
            return s1[0]+s2[0]+strin(s1[1:],s2[1:])
    else:
        return False



def count_n(s1,s2):
    global n
    n=0
    if len(s1)==len(s2) and (len(s1)>0 or len(s2)>0):
        if s1[0]==s2[0]:
            return n
        else:
            return (n+1) and count_n(s1[1:],s2[1:])
    else:
        return n

teacher's answer

def intercala( s1,s2):

if s1=='' and s2 == '':
     return ''
if s1=='' or s2 =='':
    return  False
if s1[0] == s2[0]:
    return ''

resp= intercala(s1[1:], s2[1:])
if resp != False:
    return s1[0]+s2[0]+resp
else:
    return resp

def intercala2(s1,s2):

if s1=='' and s2 == '':
     return ''
if s1=='' or s2 =='':
    return False
if s1[-1] == s2[-1]:
    return intercala2(s1[0:-1],s2[0:-1])

resp= intercala(s1[0:-1], s2[0:-1])
if resp != False:
    return resp + s1[-1]+s2[-1]
else:
    return resp

Answer 1

hey guys i found a answer, thanks for the help

def recur(s1,s2):
if len(s1)==len(s2):
    if s1[0]==s2[0]:
        return ""
    if len(s1)==1 or len(s2)==1:
        return s1[0]+s2[0]
    else:
        return s1[0]+s2[0]+recur(s1[1:],s2[1:])
else:
    return False