LeetCode - Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   /   2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:

Input: [4,9,0,5,1]
    4
   /   9   0
 / 5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        if (root == null) {
            return 0;
        }
        List<List<Integer>> resList = new ArrayList<List<Integer>>();
        List<Integer> curList = new ArrayList<Integer>();
        helper(root, resList, curList);
        return sum(resList);
    }
    
    public void helper(TreeNode node, List<List<Integer>> resList, List<Integer> curList) {
        List<Integer> list = new ArrayList<>(curList);
        list.add(node.val);
        if (node.left == null && node.right == null) {
            resList.add(list);
        }
        if (node.left != null) {
            helper(node.left, resList, list);
        }
        if (node.right != null) {
            helper(node.right, resList, list);
        }
    }
    
    public int sum(List<List<Integer>> resList) {
        int sum = 0;
        for (List<Integer> aList : resList) {
            int cur = 0;
            int size = aList.size();
            for (int i= size; i > 0; i--) {
                cur = cur + aList.get(size - i) * (int)Math.pow(10, i-1);
            }
            sum = sum + cur;
        }
        return sum;
    }

}