题解 | 二叉树中和为某一值的路径(二)

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void findPath(TreeNode* cur, int left){
        if(cur == nullptr) return ;
        path.push_back(cur->val);
        int tmp = cur->val;
        left = left - tmp;
        if(left == 0 && cur->left == nullptr && cur->right == nullptr) res.push_back(path);
        findPath(cur->left,left);
        findPath(cur->right,left);
        path.pop_back();
        
    }
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param target int整型 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > FindPath(TreeNode* root, int target) {
        res.clear();
        path.clear();
        findPath(root, target);
        return res;
    }
};