ツリーの高さを求める

Question

SQLを使用して、最後の葉までツリー構造に従いたいと思います。 Tree = 1、Knot = 3、Tree = 1、Knot = NULL（累積確率）、Tree = 2、Knot = 4;ツリー= 2、ノット= 5、ツリー= 2、ノットはヌル（累積確率）。ツリー= 3、ノット= 6、ツリー= 3、ノットはヌル（累積確率）、ツリー= 4、ノットはヌル（累積確率）。ツリー= 5、結び目はヌル（累積確率）です。ツリー= 6、結び目はヌル（累積確率）です。

Answer 1

次のようなデータを整理した場合、あなたは木、葉およびパスに関する詳細な情報を必要としない場合は、あなただけnodevalには等しい値treenodeがないかどうか確認することができ、次の

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; 

を行うことができますこのように、あなたが話している確率を仮定し

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

は、各ブランチの等しい確率です：

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; 

出力

：

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