Path Sum III

给定一个定值和一个二叉树，要求返回路径之和等于定值的条数，可以不从头节点开始

struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;

     TreeNode(int x) : val(x), left(NULL), right(NULL) {} };

参考discuss

class Solution {public:    int help(TreeNode* root, int sum, unordered_map<int, int>& store, int pre) {        if (!root) return 0;        root->val += pre;        int res = (root->val == sum) + (store.count(root->val - sum) ? store[root->val - sum] : 0);        store[root->val]++;        res += help(root->left, sum, store, root->val) + help(root->right, sum, store, root->val);        store[root->val]--;        return res;    }    int pathSum(TreeNode* root, int sum) {        unordered_map<int, int> store;        return help(root, sum, store, 0);    }};

用关联容器记录从根节点开始到某个节点的权值之和，当有这样的路径出现时，用关联容器的关键值减去定值sum，差值必然会出现在关联容器中