1つのオプションは次のとおりです：
最初のサンプルデータ：
SQL> with
2 -- sample data
3 place (id, parent_id, name, continent) as
4 (select 11, null, 'USA' , 'America' from dual union all
5 select 22, 11 , 'New York' , 'America' from dual union all
6 select 33, 22 , 'Manhattan', 'America' from dual union all
7 select 44, null, 'Brasil' , 'America' from dual union all
8 select 55, 44 , 'Rio' , 'America' from dual union all
9 select 66, null, 'France' , 'Europe' from dual union all
10 select 77, 66 , 'Paris' , 'Europe' from dual union all
11 select 88, 66 , 'Nice' , 'Europe' from dual
12 ),
13 member (id, place_id) as
14 (select 1, 22 from dual union all
15 select 2, 77 from dual union all
16 select 3, 33 from dual union all
17 select 4, 22 from dual union all
18 select 5, 55 from dual union all
19 select 6, 55 from dual union all
20 select 7, 88 from dual union all
21 select 8, 88 from dual union all
22 select 9, 88 from dual union all
23 select 10, 22 from dual
24 ),
次に、いくつかのCTE（コメントを参照）：
25 -- naively, count members of leaf nodes
26 naive as
27 (select m.place_id, count(*) cnt
28 from member m
29 group by m.place_id
30 ),
31 -- set root parent node to each row
32 cbr as
33 (select connect_by_root p.id as tpid,
34 p.id, p.parent_id, n.cnt
35 from place p left join naive n on p.id = n.place_id
36 connect by prior p.id = p.parent_id
37 start with p.parent_id is null
38 order by p.id
39 ),
40 -- how many members does each root node have?
41 sumtpid as
42 (select c.tpid, sum(c.cnt) cnt
43 from cbr c
44 group by c.tpid
45 )
46 -- join CBR + SUMTPID + PLACE for the final result
47 select c.id, c.parent_id, nvl(c.cnt, s.cnt) member_count,
48 p.name place_name,
49 p.continent
50 from cbr c join sumtpid s on s.tpid = c.tpid
51 join place p on p.id = c.id
52 order by c.id;
その結果：
ID PARENT_ID MEMBER_COUNT PLACE_NAM CONTINE
---------- ---------- ------------ --------- -------
11 4 USA America
22 11 3 New York America
33 22 1 Manhattan America
44 2 Brasil America
55 44 2 Rio America
66 4 France Europe
77 66 1 Paris Europe
88 66 3 Nice Europe
8 rows selected.
SQL>