Binary Tree Path Sum

Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

Example

Given a binary tree, and target = 5:

     1
    / \
   2   4
  / \
 2   3

return

[
  [1, 2, 2],
  [1, 4]
]

 /**
  * Definition of TreeNode:
  * public class TreeNode {
  *     public int val;
  *     public TreeNode left, right;
  *     public TreeNode(int val) {
  *         this.val = val;
  *         this.left = this.right = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @param root the root of binary tree
      * @param target an integer
      * @return all valid paths
      */
     public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
         // Write your code here
         List<List<Integer>> res = new ArrayList<List<Integer>>();
         if (root == null){
             return res;
         }
         List<Integer> path = new ArrayList<Integer>();
         helper(res, path, root, target);
         return res;
     }

     private void helper(List<List<Integer>> res, List<Integer> path, TreeNode root, int target){
         if (root.left == null && root.right == null){
             if (root.val == target) {
                 path.add(root.val);
                 res.add(new ArrayList<Integer>(path));
                 path.remove(path.size() - 1);
             }
             return;
         }
         path.add(root.val);
         if (root.left != null) {
             helper(res, path, root.left, target - root.val);
         }
         if (root.right != null) {
             helper(res, path, root.right, target - root.val);
         }
         path.remove(path.size() - 1);
         return;
     }
 }