题解 | #二叉树根节点到叶子节点和为指定值的路径#

递归

/**
 * struct TreeNode {
 *    int val;
 *    struct TreeNode *left;
 *    struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型vector<vector<>>
     */
    vector<vector<int>> res;
    vector<int> vec;
    int target;
    void helper(TreeNode* root, int sum){
        if(root == NULL) return;
        sum += root -> val;
        vec.push_back(root -> val);
        if(sum == target){
            if(root -> left == NULL && root -> right == NULL) res.push_back(vec);
        }//这里注意节点的值可能为负数，所以不能提前终止！
        helper(root -> left, sum);
        helper(root -> right, sum);
        vec.pop_back();
    }
    vector<vector<int> > pathSum(TreeNode* root, int sum) {
        // write code here
        target = sum;
        helper(root, 0);
        return res;
    }
};