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.

     ListNode * handler(int sum, int * carry, ListNode * result, ListNode * * new_head){
         if(result == NULL){
             result = new ListNode(sum % );
             (*new_head) = result;
         }else{
             result->next = new ListNode(sum % );
             result = result->next;
         }

         (*carry) = sum / ;
         return result;
     }
     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
         ListNode * new_head = NULL;
         ListNode * result = NULL;
         int carry = ;
         while(l1 != NULL || l2 != NULL){
             if(l1 == NULL){
                 int sum = l2->val + carry;
                 result = handler(sum, &carry, result, &new_head);
                 l2 = l2->next;
                 continue;
             }

             if(l2 == NULL){
                 int sum = l1->val + carry;
                 result = handler(sum, &carry, result, &new_head);
                 l1 = l1->next;
                 continue;
             }

             int sum = l1->val + l2->val + carry;
             result = handler(sum, &carry, result, &new_head);
             l1 = l1->next;
             l2 = l2->next;
         }

         if(carry !=  )
             result->next = new ListNode(carry);

         return new_head;
     }