112. 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.

由题意得，是要求出二叉树中是否存在一条路径，使得路径上的结点值之和等于给定的sum值。其中路径的定义是从根结点到叶子结点。一开始的想法是，从根开始，每次用sum减去结点的值，作为下一个结点的sum值，然后一直递归到最底层，所以有代码：

if (!root) {return (sum == 0);}return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);

然后如上面的二叉树是可以解决的。但是这样做遇到了问题，假如遇到了为[1,2]的二叉树，那么就会出错。因为在根结点往右结点走时，会判断为一条路径。但是根是有左结点的，所以不是叶子结点，所以是错误的，所以改正后需要判断是否是叶子结点。代码如下：

Code:(LeetCode运行9ms)

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool hasPathSum(TreeNode* root, int sum) {        if (!root) {            return false;        }                return hasPath(root, sum, false);    }        bool hasPath(TreeNode* root, int sum, bool hasbrother) {        if (!root) {            if (!hasbrother) {                return (sum == 0);            }            return false;        }        return hasPath(root -> left, sum - root -> val, root -> right != NULL) || hasPath(root -> right, sum - root -> val, root -> left != NULL);    }};