Я успешно смоделировал оператор Oracle CONNECT BY в SQL Server 2008, выполнив эти 2 предыдущих ответаВот а такжеВот и настраиваюсь, чтобы получить результаты, которые мне нужны. Но как мне это сделать в LINQ?

Вот пример того, что я делаю, используя фиктивную базу данных:

CREATE TABLE Employee(
 EmployeeID INT IDENTITY(1,1) PRIMARY KEY,
 Department INT NOT NULL,
 EmployeeName VARCHAR(40) NOT NULL,
 PeckingOrder INT NOT NULL,
 HigherDepartment INT NULL)

INSERT INTO Employee (Department,EmployeeName,PeckingOrder,HigherDepartment)
VALUES (1,'Bart',1,NULL),(2,'Homer',1,1),(2,'Marge',2,NULL),
       (3,'Lisa',1,2),(3,'Maggie',2,2),(3,'Santas Helper',3,1)

EmployeeID Department EmployeeName PeckingOrder HigherDepartment
    1            1     Bart            1             NULL
    2            2     Homer           1              1 
    3            2     Marge           2             NULL
    4            3     Lisa            1              2
    5            3     Maggie          2              2
    6            3     Santas Helper   3              1

и это SQL используется для возврата heirachy:

WITH n(level, PeckingOrder, Department, EmployeeName, HigherDepartment) AS 
    (SELECT 1, PeckingOrder, Department, EmployeeName, HigherDepartment
    FROM Test.dbo.Employee
    WHERE Department = 3
        UNION ALL
   SELECT n.level + 1, nplus1.PeckingOrder, nplus1.Department, nplus1.EmployeeName, nplus1.HigherDepartment 
   FROM Test.dbo.Employee as nplus1
   JOIN n ON n.HigherDepartment = nplus1.Department)
SELECT MAX(level) AS level, PeckingOrder, Department, EmployeeName, HigherDepartment   
FROM n
GROUP BY PeckingOrder, Department, EmployeeName, HigherDepartment
ORDER BY MAX(level) DESC, PeckingOrder ASC

level PeckingOrder Department EmployeeName HigherDepartment
  3         1           1           Bart             NULL
  2         1           2           Homer              1
  2         2           2           Marge             NULL
  1         1           3           Lisa               2
  1         2           3           Maggie             2
  1         3           3           Santas Helper      1