human
Solve the following python programming problem given the following code:
def datadiff(data1, data2, branch, p = True) : 
    def iterator(data1, data2) : 
        if (isinstance (data1, (list,)) and isinstance (data2, (list,))) : 
            return range (max (len (data1), len (data2))) 
else : 
            if (isinstance (data1, (dict,)) and isinstance (data2, (dict,))) : 
                return (set (data1.keys ()) | set (data2.keys ())) 
else : 
                raise TypeError 
diff = [] 
    if (data1 != data2) : 
        try : 
            it = iterator (data1, data2) 
except Exception : 
            return [(branch, it, data2)] 
for k in it : 
            newbranch = branch [:] 
            newbranch.append (k) 
            if p : 
                print (('Comparing branch: %s' % newbranch)) 
try : 
                try : 
                    v1 = data1 [k] 
except (KeyError, IndexError) : 
                    if p : 
                        print (('data1 seems does not have key = %s' % k)) 
diff.append ((newbranch, None, data2 [k])) 
                    continue 
try : 
                    v2 = data2 [k] 
except (KeyError, IndexError) : 
                    if p : 
                        print (('data2 seems does not have key = %s' % k)) 
diff.append ((newbranch, data1 [k], None)) 
                    continue 
except Exception : 
                if p : 
                    print (('data1 and data2 cannot be compared on branch: %s' % newbranch)) 
return diff.append ((newbranch, data1, data2)) 
else : 
                if (v1 != v2) : 
                    if p : 
                        print (('data1 and data2 do not match each other on branch: %s' % newbranch)) 
                        print (('v1 = %s' % v1)) 
                        print (('v2 = %s' % v2)) 
diff.extend (datadiff (v1, v2, newbranch)) 
return diff 

    
    

    Fix the buggy line: return [(branch, it, data2)]
assistant
#Here is Python program to fix the buggy line: return [(branch, it, data2)]
.

return [(branch, data1, data2)]