Add two numbers [LeetCode]

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Summary: Be careful about the last carry.

 1     ListNode * handler(int sum, int * carry, ListNode * result, ListNode * * new_head){

 2         if(result == NULL){

 3             result = new ListNode(sum % 10);

 4             (*new_head) = result;

 5         }else{

 6             result->next = new ListNode(sum % 10);

 7             result = result->next;

 8         }

 9             

10         (*carry) = sum / 10; 

11         return result;

12     }

13     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

14         ListNode * new_head = NULL;

15         ListNode * result = NULL;

16         int carry = 0;

17         while(l1 != NULL || l2 != NULL){

18             if(l1 == NULL){

19                 int sum = l2->val + carry;

20                 result = handler(sum, &carry, result, &new_head);

21                 l2 = l2->next;

22                 continue;

23             }

24             

25             if(l2 == NULL){

26                 int sum = l1->val + carry;

27                 result = handler(sum, &carry, result, &new_head);

28                 l1 = l1->next;

29                 continue;

30             }

31             

32             int sum = l1->val + l2->val + carry;

33             result = handler(sum, &carry, result, &new_head);

34             l1 = l1->next;

35             l2 = l2->next;

36         }

37         

38         if(carry != 0 )

39             result->next = new ListNode(carry);

40             

41         return new_head;

42     }