<p>You are given two <strong>non-empty</strong> linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/09/sumii-linked-list.jpg" style="width: 523px; height: 342px;" />
<pre>
<strong>Input:</strong> l1 = [7,2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [7,8,0,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> l1 = [2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [8,0,7]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> l1 = [0], l2 = [0]
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in each linked list is in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
	<li>It is guaranteed that the list represents a number that does not have leading zeros.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong>&nbsp;Could you solve it without reversing the input lists?</p>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
private:
    int dfs(stack<ListNode *> &st, ListNode *curr) {
        if (!curr)
            return 0;
        
        int sum = curr->val + dfs(st, curr->next);
        if (st.size()) {
            sum += st.top()->val;
            st.pop();
        }
        
        curr->val = sum%10;
        
        return sum/10;
    }
    
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<ListNode *> st;
        
        ListNode *t = l2;
        while (t) {
            st.push(t);
            t = t->next;
        }
        
        int carry = dfs(st, l1);
        if (st.size()) {
            st.top()->next = l1;
            while (st.size() && carry) {
                t = st.top();
                st.pop();
            
                int v = t->val + carry;
                t->val = v%10;
                carry = v/10;
            }
            
            l1 = l2;
        }
        
        if (carry && st.empty())
            return new ListNode(1, l1);
        
        return l1;
    }
};
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