Ok, I am making a lot of assumptions (INT instead of VARCHAR(50) being one of them) with this answer, so feel free to correct me if needed. The problem with option B is that it introduces a new join to relate Users to Alerts without any real added benefit. If joining on the UserID, it is best to index the UserID, so you can utilize seeks for your joins.
For Option A, UserID will be the clustering key (index key for the clustered index) on the Users table. UserID will be a nonclustered index key on Alerts table. This will cost 16 bytes per Alert.
For Option B, UserID will be the clustering key on the Users table. UserId will probably be the clustering key in UserMap too, to make joining more efficient. UserKey (assuming this is an INT) would then be a nonclustered index key on the Alerts table. This will cost 4 bytes per Alert. And 20 bytes per UserMap.
Looking at the big picture, one relationship, for Option A, costs 16 bytes of storage, and involves 1 join operation. Whereas, one relationship, for Option B, costs 24 bytes of storage, and involves 2 join operations.
Furthermore, there are a possibility of 340,282,366,920,938,000,000,000,000,000,000,000,000 uniqueidentifiers and only 4,294,967,296 INTs. Implementing a uniqueidentifier to INT map for a this type of relationship could cause unexpected results when you start reusing INTs.
The only reason for creating this type map table, is if you plan on creating a Many to Many relationship between Users and Alerts.
Taking all of this into consideration, I would recommend Option A.
I hope this helps,
Matt
OK, enough brain cells are dead.
SQL Fiddle
WITH cte AS
(
SELECT
[ICFilterID],
[ParentID],
[FilterDesc],
[Active],
CAST(0 AS varbinary(max)) AS Level
FROM [dbo].[ICFilters]
WHERE [ParentID] = 0
UNION ALL
SELECT
i.[ICFilterID],
i.[ParentID],
i.[FilterDesc],
i.[Active],
Level + CAST(i.[ICFilterID] AS varbinary(max)) AS Level
FROM [dbo].[ICFilters] i
INNER JOIN cte c
ON c.[ICFilterID] = i.[ParentID]
)
SELECT
[ICFilterID],
[ParentID],
[FilterDesc],
[Active]
FROM cte
ORDER BY [Level];
Best Answer
You can do something like this: