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/**
 * 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]
    ]
 */
 
 public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(root != null){
            ArrayList<Integer> path = new ArrayList<Integer>();
            pathSumHelper(root, sum, res, path);
        }
        return res;
    }
    
    public void pathSumHelper(TreeNode root, int sum, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path){
        if(root == null)
            return;
        if(root.left == null && root.right == null && root.val == sum){
            // path found.
            ArrayList<Integer> apath = new ArrayList<Integer>(path);
            apath.add(root.val);
            res.add(apath);
        }else{
            // walk into next level
            path.add(root.val);
            if(root.left != null)
                pathSumHelper(root.left, sum - root.val, res, path);
            if(root.right != null)
                pathSumHelper(root.right, sum - root.val, res, path);
            // recover the global variable. backtracking. 
            path.remove(path.size() - 1);
        }
    }
}

/**
 * Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
 * For example:
 * Given the below binary tree and sum = 22,
 *            5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1
 * return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
 */
 
 /**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null)
            return false;
        if(root.left == null && root.right == null && root.val == sum)
            return true;
        else
            return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}