How to get the Managers active descendants count - sql

Question

I need to create a query that display the Employee, Manager, Active, and the active descendants under the manager's hierarchy. Anyone can help me on this?
Results:
/* EmployeeId ManagerId IsActive NoofDescendantsActive
1 -1 0 4
2 1 1 3
3 1 0 2
4 2 1 2
6 2 0 0
7 3 1 0
8 3 1 0
9 4 1 1
10 9 1 0
11 9 0 0
*/
create table Person
(
EmployeeId int,
ManagerId int,
IsActive bit
);
insert into Person(EmployeeId,ManagerId,IsActive) values
(1, -1, 0),
(2, 1, 1),
(3, 1, 0),
(4, 2, 1),
(6, 2, 0),
(7, 3, 1),
(8, 3, 1),
(9, 4, 1),
(10, 9, 1),
(11, 9, 0);

Answer 1

The solution of Giorgos has a flaw. If employee #6 is made active, manager #1 should increase by 1 only. However, in his solution, it increased by 2.
EmployeeId NoofDescendantsActive
1 6
2 4
3 2
4 2
9 1
Here's a solution that works:
with Managers as
(
select
RootManagerId = EmployeeId, IsActive
from dbo.Person
where EmployeeId in (select ManagerId from dbo.Person where IsActive = 1)
),
A as -- Manager with employees
(
select m.RootManagerId, AnchorManagerId = m.RootManagerId,
ManagerId = convert(int, null)
from Managers m
union all
select m.RootManagerId, AnchorManagerId = d.EmployeeId,
d.ManagerId
from A m
-- collect descendants
join dbo.Person d on m.AnchorManagerId = d.ManagerId and d.IsActive = 1
),
Solution as
(
select RootManagerId, count(*) as Descendants
from A
where ManagerId is not null
group by RootManagerId
)
-- select * from A order by RootManagerId; -- Uncomment to see how things works
select * from Solution;
Result:
RootManagerId Descendants
1 4
2 3
3 2
4 2
9 1
Here's the result if employee #6 is made active:
RootManagerId Descendants
1 5
2 4
3 2
4 2
9 1

Answer 2

I think you can get the desired result using the following recursive CTE:
;WITH Descendants_CTE AS (
-- Anchor member: Get leaf nodes first
SELECT p1.EmployeeId, p1.ManagerId, p1.IsActive, 0 AS level
FROM Person AS p1
WHERE NOT EXISTS (SELECT 1
FROM Person AS p2
WHERE p2.ManagerId = p1.EmployeeId)
UNION ALL
-- Recursive member: Get nodes of next level, keep track of
-- the number of active nodes so far
SELECT p.EmployeeId, p.ManagerId, p.IsActive, level = level + 1
FROM Person AS p
INNER JOIN Descendants_CTE AS d ON p.EmployeeId = d.ManagerId
WHERE d.IsActive = 1
)
SELECT EmployeeId, SUM(level) AS NoofDescendantsActive
FROM Descendants_CTE
WHERE level > 0
GROUP BY EmployeeId
The recursion traverses the tree 'bottom-up', adding up the number of 'active' nodes met so far. If a non-active node is met traversal is terminated. Using a GROUP BY we can get the total number of active nodes beneath each non-leaf node. If you want to get the additional fields as well, then you have to JOIN the above query to the original table.
Hint: To better understand how the algorithm of the query actually works you can construct a graph of the tree on a piece of paper. Highlight each active node using a colored marker.
Demo here