Path Sum

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.

给定了一颗二叉树和一个整数sum，要求我们查找树中是否存在一条从根到叶子节点的和为sum的路径。我们从根节点开始搜索，用DFS，当遍历到一个节点时，从根节点到当前节点的和正好为sum，这时我们还要查看当前节点是否为叶子节点，如果为叶子节点就返回true，否则继续搜索。代码如下：

/**
 * Definition for a binary tree node.
 * 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(sum == root.val && root.left == null && root.right == null) return true;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}