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 {
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) {
        // write code here
        vector<int> path;
        vector<vector<int> > paths;
        if( root == nullptr ){
            return paths;
        }
        func( root,target,path,paths );
        return paths;
    }
    void func(TreeNode* root, int target, vector<int> &q,  vector<vector<int> > &paths){
        target -= root->val;
        q.push_back(root->val);
        if( target == 0 ){
            if( root->left == nullptr && root->right == nullptr ){
                paths.push_back(q);
            }
        }else{
            if( root->left != nullptr ) func( root->left,target,q,paths );
            if( root->right != nullptr ) func( root->right,target,q,paths );
        }
        q.pop_back();

        }
};