Leet Code OJ 112. Path Sum [Difficulty: Easy]

题目:
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,

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

翻译：
给定一个二叉树和一个和sum，检测这个二叉树是否存在一个从根节点到叶子节点的路径，每个节点的和加起来是sum。

分析：
这道题目实现起来并不复杂，但有一些特殊取值的定义，题目说的不是很清楚，比如是空树的时候，返回什么，根据OJ的结果，空树是直接返回false的。还有一个可能出错的地方是在叶子节点的判断上，例如”[1,2] sum=1”这个输入，”1”这个节点有左子树”2”而没有右子树，而计算到”1” 这个节点时值刚刚是sum的1，笔者的第一版程序返回了true，但实际上这个节点并不是一个叶子节点。

代码：

/*** 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;}else{if(root.left==null&&root.right==null&&sum==root.val){return true;}if(hasPathSum(root.left, sum-root.val)){return true;}if(hasPathSum(root.right, sum-root.val)){return true;}}return false;}
}

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