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 % 10);
            (*new_head) = result;
        }else{
            result->next = new ListNode(sum % 10);
            result = result->next;
        }
            
        (*carry) = sum / 10; 
        return result;
    }
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode * new_head = NULL;
        ListNode * result = NULL;
        int carry = 0;
        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 != 0 )
            result->next = new ListNode(carry);
            
        return new_head;
    }