DFS(深度优先搜索)--路劲总和

利用深度优先搜索遍历二叉树，每次减去对应的节点值。

题解：

/**
 * 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 sum == root.val;
        }
        return hasPathSum(root.left,sum - root.val) || hasPathSum(root.right,sum - root.val); 
    }
}