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.

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      1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution：Top-down

1. 每次更新newSum = sum - currentValue，传入左右子节点的recursion函数。
2. 如果是leaf node，且返回root.val == newSum???
3. 如果不是leaf node，则迭代得到左右子树是否能找到newSum，任一找到，就是true
   return hasPathSum (root.left, newSum) || hasPathSum (root.right, newSum);

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        
        if (root.left == null && root.right == null) {
            return root.val == sum;
        }
        
        int newSum = sum - root.val;
        return hasPathSum (root.left, newSum) || hasPathSum (root.right, newSum);
    }
}