Path Sum-二叉树相加之和

链接：https://leetcode.com/problems/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.

public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if (root == null) return false;        if(root.left==null&&root.right==null&&sum-root.val==0){            return true;        }        return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);    }}

运用递归的思想，虽然代码很简单，但是就是想不到这样做，受教了。主要是运用了递归的思想。首先判断是否为空，空则返回false，

否则进行下步判断，当前值是否有根节点，有则进行继续递归；若无则减去当前值的前提下，为零，那么就可以返回true。