Вот пример только для одного идентификатора пользователя. Вы можете использовать proc, чтобы зациклить все.
CREATE TABLE a_lnk
(user_id VARCHAR2(5),
parent_id VARCHAR2(5));
CREATE TABLE b_perm
(perm VARCHAR2(5),
user_id VARCHAR2(5));
INSERT INTO a_lnk
SELECT 1, NULL
FROM DUAL;
INSERT INTO a_lnk
SELECT 2, 1
FROM DUAL;
INSERT INTO a_lnk
SELECT 3, 1
FROM DUAL;
INSERT INTO a_lnk
SELECT 4, 3
FROM DUAL;
INSERT INTO b_perm
SELECT 'A', 1
FROM DUAL;
INSERT INTO b_perm
SELECT 'B', 3
FROM DUAL;
-- example for just for user id = 1
--
SELECT c.user_id, c.perm
FROM b_perm c,
(SELECT parent_id, user_id
FROM a_lnk
START WITH parent_id = 1
CONNECT BY PRIOR user_id = parent_id
UNION
SELECT parent_id, user_id
FROM a_lnk
START WITH parent_id IS NULL
CONNECT BY PRIOR user_id = parent_id) d
WHERE c.user_id = d.user_id
UNION
SELECT d.user_id, c.perm
FROM b_perm c,
(SELECT parent_id, user_id
FROM a_lnk
START WITH parent_id = 1
CONNECT BY PRIOR user_id = parent_id
UNION
SELECT parent_id, user_id
FROM a_lnk
START WITH parent_id IS NULL
CONNECT BY PRIOR user_id = parent_id) d
WHERE c.user_id = d.parent_id;