--给个现成的
SET NOCOUNT ON;
USE tempdb;
GO
IF OBJECT_ID('dbo.Employees') IS NOT NULL
  DROP TABLE dbo.Employees;
GO
CREATE TABLE dbo.Employees
(
  empid   INT         NOT NULL PRIMARY KEY,
  mgrid   INT         NULL    ,
  empname VARCHAR(25) NOT NULL,
  salary  MONEY       NOT NULL,
  CHECK (empid <> mgrid)
);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(1, NULL, 'David', $10000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(2, 1, 'Eitan', $7000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(3, 1, 'Ina', $7500.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(4, 2, 'Seraph', $5000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(5, 2, 'Jiru', $5500.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(6, 2, 'Steve', $4500.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(7, 3, 'Aaron', $5000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(8, 5, 'Lilach', $3500.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(9, 7, 'Rita', $3000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(10, 5, 'Sean', $3000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(11, 7, 'Gabriel', $3000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(12, 9, 'Emilia' , $2000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(13, 9, 'Michael', $2000.00);
INSERT INTO dbo.Employees(empid, mgrid, empname, salary)
  VALUES(14, 9, 'Didi', $1500.00);
  DECLARE @root AS INT;
SET @root = 3;
WITH SubsCTE
AS
(
  -- Anchor member returns root node
  SELECT empid, empname,salary , 0 AS lvl
  FROM dbo.Employees
  WHERE empid = @root
  UNION ALL
  -- Recursive member returns next level of children
  SELECT C.empid, C.empname,c.salary , P.lvl + 1
  FROM SubsCTE AS P
    JOIN dbo.Employees AS C
      ON C.mgrid = P.empid
)
SELECT SUM(salary) FROM SubsCTE;