二叉树中和为某一数值的路径

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

    void dfs(vector<vector<int>> &res, TreeNode* root, int number, vector<int>& trace){
        trace.push_back(root->val); // 记录当前节点的值
        if(root->left == nullptr && root->right == nullptr){
            if(number == root->val){   // 叶节点且路径之和为expectNumber，结束递归
                res.push_back(trace);
            }
        }else{
            if(root->left){
                dfs(res, root->left, number-root->val, trace);
            }
            if(root->right){
                dfs(res, root->right, number-root->val, trace);
            }
        }
        trace.pop_back();
    }

    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        vector<vector<int>> res;
        vector<int> trace;
        if(root != nullptr){
            dfs(res, root, expectNumber, trace);
            return res;
        }
        return res;
    }
};