给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

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