oracle 递归 求和

create table test( NAME varchar2(20), P_NAME varchar2(20), NUM int);  
insert into test values('1','-1',3 );
insert into test values('11','1',5 );
insert into test values('111','11',6 );
insert into test values('1111','111',9 );
insert into test values('2','-1',4 );
insert into test values('21','2',7 );
insert into test values('211','21',8 );
COMMIT;
SELECT NAME, P_NAME, SUM(NUM) num
  FROM (
        SELECT NUM, CONNECT_BY_ROOT NAME NAME, CONNECT_BY_ROOT P_NAME P_NAME
          FROM TEST S
        CONNECT BY PRIOR NAME = P_NAME
         --START WITH P_NAME = '-1'
       )
 GROUP BY NAME, P_NAME order by name;