二叉树中和为某一值的路径

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

class Solution {
public:
    vector<vector<int>>res;
    vector<int> path;
    void DFS(TreeNode*root,int num)
    {
        if(root==NULL) return;
        path.push_back(root->val);
        if(!root->left&&!root->right&&num==root->val)//根节点的左右子树没有，且根节点的值等于num（所求的值），把向量path放入res中即可
            res.push_back(path);
        else//左右子树存在
        {
            if(root->left)
               DFS(root->left,num-root->val);
            if(root->right)
                DFS(root->right,num-root->val);
        }
        path.pop_back();//相当于回退 ，弹出去
    }
    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        DFS(root,expectNumber);
        return res;

    }
};