SQL Get lowest level child and root node - sql

Question

I have database schema: [Id], [ParrentId], [some more tables]
I have hierarchy like:
1. a
2. aa
3. aaa_1
3. aaa_2
1. b
2. bb
1. c
2. cc
3. ccc_1
4. cccc
3. ccc_2
I want a (select * where X) => [X, lowest leve child] like:
[a, aaa_1] [a, aaa_2]; [cc, cccc] etc.
I can get lowest child with
SELECT t1.name FROM
category AS t1 LEFT JOIN category as t2
ON t1.category_id = t2.parent
WHERE t2.category_id IS NULL;
but I don't know how to join it with root node.

Answer 1

Given:
The DBMS is SQL Server;
The highest level nodes of the tree have parent = NULL;
You want all the lowest leaves for all levels of the trees, not just the roots;
You want to have all the nodes at a lowest level, not just one;
This query would do it:
WITH r ( category_id, name, root, depth )
-- finds the root relationship
AS (
SELECT category_id, name, category_id, 0
FROM category
-- WHERE parent IS NULL -- this would only look at root nodes
UNION ALL
SELECT c.category_id, c.name, r.root, r.depth + 1
FROM r
JOIN category c
ON c.parent = r.category_id
), s ( category_id, name, root, window_id )
-- finds the lowest leaves
AS (
SELECT category_id, name, root, RANK() OVER(partition by root order by depth DESC)
FROM r
)
SELECT c.name AS NodeName, s.Name AS DeepLeafName
FROM category c
JOIN s
ON c.category_id = s.root
WHERE s.window_id = 1;
Here is the result set:

Answer 2

With SQL Server, you can try this :
With CTE as
(
Select ID as Child, lev = 1
from category
where ID = X
UNION ALL
Select category.ID, CTE.lev + 1
from category
inner join CTE ON category.ParentID = CTE.Child
)
select CTE_1.Child, CTE_2.Child
from CTE as CTE_1
inner join CTE as CTE_2
where CTE_1.lev = 1 AND CTE_2.lev = (select MAX(CTE.lev) from CTE)