LeetCode OJ 之 Path Sum(求路径和）

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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.

给定一个二叉树和一个数，确定树中是否有从根到叶结点路径上的值的和与所给的值相等。

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.

/** * Definition for binary tree * 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)     {        //思路：利用sum减去当前遍历结点的值，直到叶结点，再比较判断是否符合条件        if (root == NULL)             return false;        //递归结束条件，到达叶结点，判断        if (root->left == NULL && root->right == NULL)            return sum == root->val;        //        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);    }};

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