忽略分层查询中的单个孩子（竹子部分）

Question

我有一个分层数据如下的表格。

create table tst as 
select 1 id, null parent_id from dual union all 
select 2 id, 1 parent_id from dual union all 
select 3 id, 1 parent_id from dual union all 
select 4 id, 2 parent_id from dual union all 
select 5 id, 3 parent_id from dual union all 
select 6 id, 5 parent_id from dual union all 
select 7 id, 6 parent_id from dual union all 
select 8 id, 6 parent_id from dual; 

使用CONNECT BY语句遍历层次结构是微不足道的。我有的提取要求是忽略树的简单（类似于竹的）部分，即如果一个父节点只有一个子节点，则两者都连接在一起并且连接这些ID（此规则是递归应用的）。

因此，预期的结果是

 ID PARENT_ID 
---------- ---------- 
     1    
     2,4  1 
     3,5,6  1 
     7   3,5,6 
     8   3,5,6 

UPDATE或者这也是正确答案（加入串联节点列表和重复使用原有IDS）

 ID PARENT_ID NODE_LST 
---------- ---------- --------- 
     1   1  
     4   1 2,4  
     6   1 3,5,6 
     7   6 7  
     8   6 8 

这远远我设法算该孩子并建立完整的路径，以儿童计数和ID的根...

with child_cnt as (
-- child count per parent 
select parent_id, count(*) cnt 
from tst 
where parent_id is not NULL 
group by parent_id), 
tst2 as (
select 
    ID, child_cnt.cnt, 
    tst.parent_id 
from tst left outer join child_cnt on tst.parent_id = child_cnt.parent_id), 
tst3 as (
SELECT id, parent_id, 
    sys_connect_by_path(cnt,',') child_cnt_path, 
    sys_connect_by_path(id,',') path 
FROM tst2 
    START WITH parent_id IS NULL 
    CONNECT BY parent_id = PRIOR id 
) 
select * from tst3 
; 


     ID PARENT_ID CHILD_CNT_PATH PATH  
---------- ---------- -------------- ------------ 
     1   ,    ,1   
     2   1 ,,2   ,1,2   
     4   2 ,,2,1   ,1,2,4  
     3   1 ,,2   ,1,3   
     5   3 ,,2,1   ,1,3,5  
     6   5 ,,2,1,1  ,1,3,5,6  
     7   6 ,,2,1,1,2  ,1,3,5,6,7 
     8   6 ,,2,1,1,2  ,1,3,5,6,8 

这将建议在ID 4和5上完成一个级别的跳过（一个尾随子数1），并且在ID 6上跳过2级（计数路径中的两个训练数据）。

但我认为应该有一个更简单的方法来解决这个问题。

Answer 1

这不是很优雅，但它应该工作。如果我能找出更好的方法来完成最后的部分，我会编辑。祝你好运！

with 
    d (id, parent_id, degree) as (
     select id, parent_id, count(parent_id) over (partition by parent_id) 
     from tst 
    ), 
    x (old_id, new_id) as (
     select id, ltrim(sys_connect_by_path(id, ','), ',') 
     from d 
     where connect_by_isleaf = 1 
     start with degree != 1 
     connect by parent_id = prior id 
     and  degree = 1 
    ) 
select x1.new_id as id, x2.new_id as parent_id 
from x x1 
      inner join tst 
       on tst.id  = regexp_substr(x1.new_id, '^[^,]+') 
      left outer join x x2 
       on tst.parent_id = x2.old_id 
; 

Answer 2

此查询将帮助您找到其他解决方案。

尽管可能会进一步优化或修复一些错误，但它适用于您的测试用例。

WITH nodes_to_dispose as (
    SELECT min(id) as id, 
      parent_id 
    FROM tst 
    WHERE parent_id is not null 
    GROUP BY parent_id 
    HAVING count(*) = 1) 
-- This part returns merged bamboo nodes 
SELECT nodes_to_dispose.id, 
     connect_by_root tst.parent_id as parent_id, 
     connect_by_root nodes_to_dispose.parent_id || 
       sys_connect_by_path(nodes_to_dispose.id, ',') as node_lst 
FROM nodes_to_dispose, tst 
WHERE nodes_to_dispose.parent_id = tst.id (+) 
AND connect_by_isleaf = 1 
START WITH nodes_to_dispose.parent_id not in (
    SELECT id 
    FROM nodes_to_dispose) 
CONNECT BY prior nodes_to_dispose.id = nodes_to_dispose.parent_id 
UNION 
-- This part returns all other nodes in their original form 
SELECT id, parent_id, to_char(id) as node_lst 
FROM tst 
WHERE id not in (
    SELECT parent_id 
    FROM nodes_to_dispose 
    UNION 
    SELECT id 
    FROM nodes_to_dispose);