SQL 查询拓扑排序

Question

我有一个有向无环图：

DROP TABLE IF EXISTS #Edges
CREATE TABLE #Edges(from_node int, to_node int);
INSERT INTO #Edges VALUES (1,2),(1,3),(1,4),(5,1);

我想列出所有节点，始终在其起始节点之前列出终止节点。例如：2、3、4、1、5。

也叫拓扑排序。如何在 SQL 中完成？

Answer 1

您可以使用递归 CTE 来计算深度。然后按深度排序：

with cte as (
      select e.from_node, e.to_node, 1 as lev
      from edges e
      where not exists (select 1 from edges e2 where e2.to_node = e.from_node)
      union all
      select e.from_node, e.to_node, lev + 1
      from cte join
           edges e
           on e.from_node = cte.to_node
     )
select *
from cte
order by lev desc;

编辑：

我注意到您的边缘列表中没有“1”。要处理此问题：

with cte as (
      select 1 as from_node, e.from_node as to_node, 1 as lev
      from edges e
      where not exists (select 1 from edges e2 where e2.to_node = e.from_node)
      union all
      select e.from_node, e.to_node, lev + 1
      from cte join
           edges e
           on e.from_node = cte.to_node
      -- where lev < 5
     )
select *
from cte
order by lev desc;

Here 是一个 db<>fiddle.

Answer 2

DROP TABLE IF EXISTS #topological_sorted
CREATE TABLE #topological_sorted(id int identity(1,1) primary key, n int);


WITH rcte(n) AS (

    SELECT e1.to_node
    FROM #Edges AS e1
    LEFT JOIN #Edges AS e2 ON e1.to_node = e2.from_node 
    WHERE e2.from_node IS NULL

    UNION ALL

    SELECT e.from_node
    FROM #Edges AS e 
    JOIN rcte ON e.to_node = rcte.n

)
INSERT INTO #topological_sorted(n)
SELECT *
FROM rcte;

SELECT * FROM #topological_sorted

节点可能会被多次列出。我们只想保留第一次出现：

DROP TABLE IF EXISTS #topological_sorted_2

SELECT *, MIN(id) OVER (PARTITION BY n) AS idm
INTO #topological_sorted_2
FROM #topological_sorted 
ORDER BY id;

SELECT * FROM #topological_sorted_2
WHERE id=idm
ORDER BY id;

SQL 查询拓扑排序

SQL query for topological sort

sql

sql-server

graph-algorithm

graph-traversal

topological-sort