113. Path Sum II

 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

My answer is:

/**
 * 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 List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        helper(root, sum, 0, new ArrayList<Integer>(), result);
        return result;
    }
    
    public void helper(TreeNode node, int sum, int currentSum, List<Integer> currentNodes, List<List<Integer>> result) {
        if (node == null) {
            return;
        }
        currentSum += node.val;
        currentNodes.add(node.val);
        if (node.left == null && node.right == null) {
            if (currentSum == sum) {
                List<Integer> foundNodes = new ArrayList<Integer>(currentNodes);
                result.add(foundNodes);
            }
            currentNodes.remove(currentNodes.size() - 1);
            return;
        }
        
        helper(node.left, sum, currentSum, currentNodes, result);
        helper(node.right, sum, currentSum, currentNodes, result);
        currentNodes.remove(currentNodes.size() - 1);
    }
}