如何使用VIEW将Postgresql ltree结构转换为嵌套集？

我想问一个问题，关于如何通过仅使用视图将PostgreSQL ltree结构转换为只有一个查询的嵌套集结构

例如，我有一个表，其中包含相互关联的数据，如下图所示：

因此，表Daclariation是

KeywordLtree(id INT PRIMARY KEY, value TEXT, path ltree);

-- And the data is:

pk |  value  |  path  |
0  |   'A'   |   ''   |
0  |   'B'   |   '1'  |
0  |   'C'   |   '2'  |
0  |   'D'   |  '1.3' |
0  |   'E'   |  '1.4' |
0  |   'F'   |  '1.5' |
0  |   'G'   |  '2.6' |
0  |   'H'   |  '2.7' |

我必须把这个表转换成这样的表：

KeywordSets(id INT PRIMARY KEY, value TEXT, lft INT, rgt INT);

其中，左边框和右边框的规则是根据嵌套规则集完成的。 我找到了在每个级别上获取顶点的方法

CREATE OR REPLACE RECURSIVE VIEW bfs (id, value, path, num_on_level, level) AS
    SELECT id, value, path, row_number() OVER (), 0 as level
    FROM KeywordLtreeSandbox WHERE path ~ '*{0}'

    UNION ALL

    SELECT C.id, C.value, C.path, row_number() OVER (PARTITION BY P.path), level + 1 
    FROM KeywordLtreeSandbox C JOIN bfs P ON P.path @> C.path 
    WHERE nlevel(C.path) = level + 1;

-- Data would be like below
id | value | path | num_on_level | level |
0  |  "A"  |  ""  |      1       |   0   |
1  |  "B"  | "1"  |      1       |   1   |
2  |  "C"  | "2"  |      2       |   1   |
3  |  "D"  |"1.3" |      1       |   2   |
4  |  "E"  |"1.4" |      2       |   2   |
5  |  "F"  |"1.5" |      3       |   2   |
6  |  "G"  |"2.6" |      1       |   2   |
7  |  "H"  |"2.7" |      2       |   2   |

但我不知道如何正确地列举它们，所以A左=1，右=16，B左=2，右=9等等

如果我需要更清楚，请让我知道

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有人能告诉我怎么做吗？

关于stepic.org上的课程-

WITH RECURSIVE bfs(id, value, path, level, cnt) AS
        (SELECT id, value, path, 0 as level,1 as cnt 
        FROM kts WHERE path ~ '*{0}'

    UNION ALL

    SELECT C.id, C.value, C.path, level + 1, 1 as cnt
    FROM kts C JOIN bfs P ON P.path @> C.path 
    WHERE nlevel(C.path) = level + 1),

qsubtrees AS (SELECT K2.path AS p, COUNT(*) AS size FROM kts K LEFT JOIN kts K2 ON K.path <@ K2.path
           GROUP BY K2.path),

result AS (SELECT id, value, path, level,
 2*(SUM(cnt) OVER (ROWS UNBOUNDED PRECEDING))-1-level AS leftn, size
       FROM bfs JOIN qsubtrees ON bfs.path=qsubtrees.p  ORDER BY path)



SELECT '#', id, value, leftn, leftn+2*size-1 AS rightn FROM result ORDER BY leftn; 

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哇，你在我的视图中添加了一些CTE。第二个用于计算子树，但是你能解释一下结果CTE是什么吗？