程式語言 - LeetCode - C++ - 437. Path Sum III

參考資訊：
https://www.cnblogs.com/grandyang/p/6007336.html

題目：


解答：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int targetSum) {
        int r = 0;
        vector<int> q;

        dfs(root, targetSum, 0, q, r);
        return r;
    }

    int dfs(TreeNode *n, int targetSum, long curSum, vector<int> &q, int &r) {
        if (!n) {
            return -1;
        }

        curSum += n->val;
        if (curSum == targetSum) {
            r += 1;
        }

        long tmp = curSum;
        for (int i = 0; i < q.size(); i++) {
            tmp -= q[i];
            if (tmp == targetSum) {
                r += 1;
            }
        }

        q.push_back(n->val);
        dfs(n->left, targetSum, curSum, q, r);
        dfs(n->right, targetSum, curSum, q, r);
        q.pop_back();
        return 0;
    }
};