Path Sum III（C++路徑總和 III）

原網址：https://leetcode.com/problems/path-sum-iii/discuss/91877/C%2B%2B-5-Line-Body-Code-DFS-Solution

解題思路：

（1）總的數目是以當前爲根的數目加上以左孩子和右孩子分別爲根的數目

（2）作者的雙遞歸用的確實很驚豔

/**
 * 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:
    int pathSum(TreeNode* root, int sum) {
        if(!root) return 0;
        return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
private:
    int sumUp(TreeNode* root, int pre, int& sum){
        if(!root) return 0;
        int current = pre + root->val;
        return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);
    }
};