Question

使用SQL，我想按照树结构直到最后一页。 Tree = 1且Knot = 2; Tree = 1且Knot = 3; Tree = 1且Knot = NULL（累积概率）; Tree = 2且Knot = 4; Tree = 2且Knot = 5; Tree = 2且Knot为null（累积概率）; Tree = 3且Knot = 6; Tree = 3且Knot为null（累积概率）; Tree = 4且Knot为null（累积概率）; Tree = 5且Knot为null（累积概率）; Tree = 6且Knot为null（累积概率）;

Answer 1

如果您组织这样的数据，您可以执行以下操作

with TREETAB as (
select '1' treenode, '2' nodeval from dual
union all
select '1' treenode, '3' nodeval from dual
union all
select '1' treenode, 'leaf1' nodeval from dual
union all
select '2' treenode, '4' nodeval from dual
union all
select '2' treenode, '5' nodeval from dual
union all
select '2' treenode, 'leaf2' nodeval from dual
union all
select '3' treenode, '4' nodeval from dual
union all
select '3' treenode, 'leaf3' nodeval from dual
union all
select '4' treenode, 'leaf4' nodeval from dual
union all
select '5' treenode, 'leaf5' nodeval from dual
union all
select '6' treenode, 'leaf6' nodeval from dual
)

select *
  from (SELECT t.*, CONNECT_BY_ISLEAF Leaf,
                SYS_CONNECT_BY_PATH(nodeval, '/') "Path", level
           FROM TREETAB t
          start with t.treenode = 1
         CONNECT BY PRIOR t.nodeval = to_char(t.treenode))
 where Leaf = 1
 order by treenode;

如果不需要有关树，叶和路径的详细信息，您可以检查treenode这样的nodeval没有相同的值

with TREETAB as
 (select '1' treenode, '2' nodeval
    from dual
  union all
  select '1' treenode, '3' nodeval
    from dual
  union all
  select '1' treenode, 'leaf1' nodeval
    from dual
  union all
  select '2' treenode, '4' nodeval
    from dual
  union all
  select '2' treenode, '5' nodeval
    from dual
  union all
  select '2' treenode, 'leaf2' nodeval
    from dual
  union all
  select '3' treenode, '4' nodeval
    from dual
  union all
  select '3' treenode, 'leaf3' nodeval
    from dual
  union all
  select '4' treenode, 'leaf4' nodeval
    from dual
  union all
  select '5' treenode, 'leaf5' nodeval
    from dual
  union all
  select '6' treenode, 'leaf6' nodeval from dual)

select *
  from TREETAB t
 where not exists
 (select t.treenode from TREETAB t2 where t2.treenode = t.nodeval)

Answer 2

假设您所谈论的概率对于每个分支是相同的概率，那么：

WITH data ( tree, knot ) AS (
  SELECT 1, 2 FROM DUAL UNION ALL
  SELECT 1, 3 FROM DUAL UNION ALL
  SELECT 1, NULL FROM DUAL UNION ALL
  SELECT 2, 4 FROM DUAL UNION ALL
  SELECT 2, 5 FROM DUAL UNION ALL
  SELECT 2, NULL FROM DUAL UNION ALL
  SELECT 3, 6 FROM DUAL UNION ALL
  SELECT 3, NULL FROM DUAL UNION ALL
  SELECT 4, NULL FROM DUAL UNION ALL
  SELECT 5, NULL FROM DUAL UNION ALL
  SELECT 6, NULL FROM DUAL 
),
probabilities ( tree, knot, probability ) AS (
  SELECT tree, knot, 1 / COUNT(1) OVER ( PARTITION BY tree ) FROM data
),
cumulative_probabilities ( tree, knot, probability, path ) AS (
  SELECT p.*, '/' || tree
  FROM   probabilities p
  WHERE  tree = 1
UNION ALL
  SELECT p.tree, p.knot, p.probability * cp.probability, cp.path || '/' || p.tree
  FROM   probabilities p
         INNER JOIN
         cumulative_probabilities cp
         ON ( p.tree = cp.knot )
)
SELECT *
FROM   cumulative_Probabilities
WHERE  knot IS NULL;

<强>输出：

TREE KNOT PROBABILITY PATH
---- ---- ----------- ------
   1      0.333333333 /1
   3      0.166666667 /1/3
   2      0.111111111 /1/2
   6      0.166666667 /1/3/6
   4      0.111111111 /1/2/4
   5      0.111111111 /1/2/5

找到树的高度

2 个答案: