Oracle 层级树（子选父，父选子）

create table TEST_TREE
(
  id VARCHAR2(32),
  name VARCHAR2(32),
  pid VARCHAR2(32),
  type VARCHAR2(32),
  lev VARCHAR2(32)
);
comment on column TEST_TREE.name
  is '名称';
comment on column TEST_TREE.pid
  is '父ID';
comment on column TEST_TREE.type
  is '种类';
comment on column TEST_TREE.lev
  is '级别';
insert into TEST_TREE (id, name, pid, type, lev)
values ('0001', '中国', null, null, '0');
insert into TEST_TREE (id, name, pid, type, lev)
values ('000101', '北京', '0001', '1', '1');
insert into TEST_TREE (id, name, pid, type, lev)
values ('000102', '湖南省', '0001', '2', '1');
insert into TEST_TREE (id, name, pid, type, lev)
values ('000103', '河南省', '0001', '2', '1');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010101', '东城区', '000101', '1', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010102', '西城区', '000101', '2', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010103', '崇文区', '000101', '4', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010201', '长沙市', '000102', '1', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010202', '岳阳市', '000102', '3', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010203', '怀化市', '000102', '1', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010301', '郑州市', '000103', '1', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010302', '开封市', '000103', '2', '2');
insert into TEST_TREE (id, name, pid, type, lev)
values ('00010303', '洛阳市', '000103', '4', '2');

create table TEST_TREE
(
  id VA