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

For 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]
]

分析： DFS backtracking... ebay面过一个题， print the path from root to a given node， 早练练这题的话那个题也不会挂了，之所以会挂主要是对Java Pass by Value的理解不到位。还有就是能够设一个全局变量boolean flag。来设置找到given node之后就不继续做recursion了

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null){
            return res;
        }
        helper(res, root, sum, new ArrayList<Integer>());
        return res;
    }
    
    private void helper(List<List<Integer>> res, TreeNode root, int sum, List<Integer> item){
        if(root == null){
            return;
        }
        if(root.left == null && root.right == null && root.val == sum){
            item.add(root.val);
            res.add(new ArrayList<Integer>(item));
            item.remove(item.size() - 1);
            return;
        }
        
        item.add(root.val);
        helper(res, root.left, sum - root.val, item);
        helper(res, root.right, sum - root.val, item);
        item.remove(item.size() - 1);
    }
}