SQL referrals view - sql

Question

I have a table with users:
id referred_by_user_id
-----------------------------------
1 null
2 1
3 1
4 2
I need to write request to get number of people referred by user in two levels.
First - direct referral (example: user 1 referred users 2 3. count for level 1 = 2)
Second - user 1 referred to users 2 and 3, user 2 referred to user 4. So count for level 2 should be 1
Result of query should be:
id referred_user_tier_one_total referred_user_tier_two_total
------------------------------------------------------------------------------
1 2 1
2 1 null
3 null null
4 null null
I figured out how to count referred_user_tier_one_total:
select
"id", referred_user_tier_one_total
from
"user"
inner join
(select
count(*) as referred_user_tier_one_total, referred_by_user_id
from
"user"
where
"user".referred_by_user_id is not null
group by
"user".referred_by_user_id) ur on "user".id = ur.referred_by_user_id
But I don't understand how to calculate referred_user_tier_two_total. Please, help
UPD:
Thanks #Stoff for SQL Server solution.
Here is the script rewritten for Postgres
WITH RECURSIVE agg AS
(
SELECT
a.ID, a.referred_by_user_id,
COUNT(b.referred_by_user_id) AS "count"
FROM
"user" a
LEFT JOIN
"user" b ON a.ID = b.referred_by_user_id
GROUP BY
a.ID, a.referred_by_user_id
)
SELECT
a.ID,
a.Count AS referred_user_tier_one_total,
CASE
WHEN SUM(b.count) IS NULL
THEN 0
ELSE SUM(b.count)
END AS referred_user_tier_two_total
FROM
agg a
LEFT JOIN
agg b ON a.ID = b.referred_by_user_id
GROUP BY
a.ID, a.Count
ORDER BY
a.ID

Answer 1

Here is a solution which works in Postgres.
We could carry on writing more levels in the same way.
create table referals(
id int,
referred_by_user_id int);
insert into referals values
(1 , null),
(2 , 1),
(3 , 1),
(4 , 2 );
select
t0.id,
count(t1.id) tier1,
count(t2.id) tier2
from
referals t0
left join referals t1
on t0.id = t1.referred_by_user_id
left join referals t2
on t1.id = t2.referred_by_user_id
group by t0.id
id | tier1 | tier2
-: | ----: | ----:
1 | 2 | 1
2 | 1 | 0
3 | 0 | 0
4 | 0 | 0
db<>fiddle here