113:Path Sum II【树】【DFS】

题目链接：click~

/*题意：给出一颗树和一个值sum，求树中从根结点到叶子结点的所有路径         且每条路径上结点的和等于sum   */  class Solution {public:    void GetPathSum(TreeNode *T, int sum, vector<int> &v, vector<vector<int>> &path) {        if(T == NULL) return;        //当前结点是叶子结点，且结点值等于sum        if(T->left == NULL && T->right == NULL && T->val == sum) {            v.push_back(T->val);            path.push_back(v);            v.pop_back();            return;        }        v.push_back(T->val);        int next_sum = sum - T->val;        GetPathSum(T->left, next_sum, v, path);        GetPathSum(T->right, next_sum, v, path);        v.pop_back();    }    vector<vector<int> > pathSum(TreeNode *root, int sum) {        vector<vector<int>> path;        vector<int> v;//存放每一个可能的路径        GetPathSum(root, sum, v, path);        return path;    }};