1. 题目

员工表：Employees

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| employee_id   | int     |
| employee_name | varchar |
| manager_id    | int     |
+---------------+---------+
employee_id 是这个表的主键。
这个表中每一行中，employee_id 表示职工的 ID，
employee_name 表示职工的名字，
manager_id 表示该职工汇报工作的直线经理。
这个公司 CEO 是 employee_id = 1 的人。

用 SQL 查询出所有直接或间接向公司 CEO 汇报工作的职工的 employee_id 。

由于公司规模较小，经理之间的间接关系不超过 3 个经理。

可以以任何顺序返回的结果，不需要去重。

查询结果示例如下：

Employees table:
+-------------+---------------+------------+
| employee_id | employee_name | manager_id |
+-------------+---------------+------------+
| 1           | Boss          | 1          |
| 3           | Alice         | 3          |
| 2           | Bob           | 1          |
| 4           | Daniel        | 2          |
| 7           | Luis          | 4          |
| 8           | Jhon          | 3          |
| 9           | Angela        | 8          |
| 77          | Robert        | 1          |
+-------------+---------------+------------+Result table:
+-------------+
| employee_id |
+-------------+
| 2           |
| 77          |
| 4           |
| 7           |
+-------------+公司 CEO 的 employee_id 是 1.
employee_id 是 2 和 77 的职员直接汇报给公司 CEO。
employee_id 是 4 的职员间接汇报给公司 CEO 4 --> 2 --> 1 。
employee_id 是 7 的职员间接汇报给公司 CEO 7 --> 4 --> 2 --> 1 。
employee_id 是 3, 8 ，9 的职员不会直接或间接的汇报给公司 CEO。 

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/all-people-report-to-the-given-manager
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2. 解题

• 一次连接

select *
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id

{"headers": ["employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id"], 
"values": [
[1, "Boss",    1, 1, "Boss", 1], 
[3, "Alice",   3, 3, "Alice", 3], 
[2, "Bob",     1, 1, "Boss", 1], 
[4, "Daniel",  2, 2, "Bob", 1], 
[7, "Luis",    4, 4, "Daniel", 2], 
[8, "John",    3, 3, "Alice", 3], 
[9, "Angela",  8, 8, "John", 3], 
[77, "Robert", 1, 1, "Boss", 1]]}

• 二次连接

select *
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id
left join Employees e3
on e2.manager_id = e3.employee_id

{"headers": ["employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id", "employee_id", "employee_name", "manager_id"], 
"values": [
[1, "Boss",    1, 1, "Boss",   1, 1, "Boss",  1], 
[3, "Alice",   3, 3, "Alice",  3, 3, "Alice", 3], 
[2, "Bob",     1, 1, "Boss",   1, 1, "Boss",  1], 
[4, "Daniel",  2, 2, "Bob",    1, 1, "Boss",  1], 
[7, "Luis",    4, 4, "Daniel", 2, 2, "Bob",   1], 
[8, "John",    3, 3, "Alice",  3, 3, "Alice", 3], 
[9, "Angela",  8, 8, "John",   3, 3, "Alice", 3], 
[77, "Robert", 1, 1, "Boss",   1, 1, "Boss",  1]]}

• 答案

# Write your MySQL query statement below
select e1.employee_id
from Employees e1 left join Employees e2
on e1.manager_id = e2.employee_id
left join Employees e3
on e2.manager_id = e3.employee_id
where e3.manager_id = 1 and e1.employee_id != 1

173 ms

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