2. Add Two Numbers

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

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

递归，每次递归时建立root的next节点，然后将root移动到root.next。
计算两个节点的和并填入root节点。每次计算时需要计算是否进位。
递归两个节点的next节点，并将carry传入。
当一个节点为null时，只递归和计算另一个节点。
当两个节点为null时，如果有carry需要将其放入新节点，如果没有则返回。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode root = new ListNode();
        addTwo(l1,l2,root,0);
        return root.next;
    }
    
    private void addTwo(ListNode l1, ListNode l2, ListNode root, int carry){
        if(l1 == null && l2 == null && carry == 0){
            return;
        }
        else if(l1 == null && l2 == null){
            root.next = new ListNode();
            root = root.next;
            root.val = carry;
            return;
        }
        else if(l1 == null){
            root.next = new ListNode();
            root = root.next;
            root.val = (l2.val + carry) % 10;
            carry = (l2.val + carry) / 10;
            addTwo(null, l2.next, root, carry);
            return;
        }
        else if(l2 == null){
            root.next = new ListNode();
            root = root.next;
            root.val = (l1.val + carry) % 10;
            carry = (l1.val + carry) / 10;
            addTwo(null, l1.next, root, carry);
            return;
        }
        root.next = new ListNode();
        root = root.next;
        root.val = (l1.val + l2.val + carry) % 10;
        carry = (l1.val + l2.val + carry) / 10;

        addTwo(l1.next, l2.next, root, carry);
    }
}