Krdytkyu

I have a non-binary tree of customer, and I need to obtain all the id in a tree for the given node.

The table is very simple just an join table with parent id and child id
This is a representation of the tree I stored in my db.

In this example if I search for node 17 I need in return 14-17
if I search for 11 I need in return 1-6-5-4-8-11-12-7-2-10-3

The order is not important I only need the id to avoid circularity when add children to a node.

I created this query.
The ancestor part works fine, I retrieve all parent nodes,
but for the descendants I have some trouble.
I'm only able to retrieve some part of the tree.
For example, with node 11 I retrieve 4-10-6-11-7-8, so all right part of the tree is missing

WITH RECURSIVE 

-- starting node(s)

starting (parent, child) AS

(

  SELECT t.parent, t.child

  FROM public.customerincustomer AS t

  WHERE t.child = :node or t.parent = :node

)

,

ancestors (parent, child) AS

(

  SELECT t.parent, t.child 

  FROM public.customerincustomer AS t 

  WHERE t.parent IN (SELECT parent FROM starting)

  UNION ALL

  SELECT t.parent, t.child 

  FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent

),

descendants (parent, child) AS

(

  SELECT t.parent, t.child 

  FROM public.customerincustomer AS t 

  WHERE t.parent IN (SELECT parent FROM starting) or t.child in (select child from starting) 

  UNION ALL

  SELECT t.parent, t.child 

  FROM public.customerincustomer AS t JOIN ancestors AS a ON t.parent = a.child

)



table ancestors

union all 

table descendants

UPDATE

I see that many example include in the tree table also the root in form (root_id, null)
in my case i don't have this record.
For example taking the smallest tree 14->17, in my table i have only one record
parent, child
14 17

A very primitive implementation:

It basically divides the problem into two subproblems:

• First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.


• Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.


• Note that the query will work even if the input node is a root with has no children.


• It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).

The query:

WITH RECURSIVE 

ancestors (parent) AS

(

  SELECT :node                      -- start with the given node

  UNION ALL

  SELECT t.parent                   -- and find all its ancestors

  FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent

),

descendants (customer) AS

(

  SELECT parent AS customer       -- now start with all the ancestors

  FROM ancestors

  UNION 

  SELECT t.child                  -- and find all their descendants

  FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer

)

SELECT customer

FROM descendants ;

This function returns the parent level of node_id:

_{There is a 'level' row due there isn't a row (id, null) for parent row.}

CREATE FUNCTION get_parent(node_id int)

RETURNS integer AS

$$

  WITH RECURSIVE get_parent AS

  (

      SELECT 

          t1.id, 

          t1.parent_id, 

          t1.name, 

          0 AS level

      FROM 

          tree t1

      WHERE 

          t1.id = node_id

      UNION ALL

      SELECT 

          t2.id, 

          t2.parent_id, 

          t2.name,

          level+1

      FROM 

          tree t2 

      INNER JOIN 

          get_parent ON get_parent.parent_id = t2.id

  )

  SELECT

      id

  FROM

      get_parent

  ORDER BY

      level DESC

  LIMIT 1 ;

$$

LANGUAGE SQL;

select get_parent(7);

| get_parent |

| ---------: |

|          6 |

Now, next query returns the whole tree structure based on a parent node.

WITH RECURSIVE childs AS

(

    SELECT 

        t1.id, 

        t1.parent_id, 

        t1.name

    FROM 

        tree t1

    WHERE 

        t1.id = get_parent(7)

    UNION ALL

    SELECT 

        t2.id, 

        t2.parent_id, 

        t2.name

    FROM 

        tree t2 

    INNER JOIN 

        childs ON childs.id = t2.parent_id

)

SELECT

    id,

    parent_id,

    name

 FROM

    childs;

id | parent_id | name   

-: | --------: | :------

 6 |         1 | Node 6 

 4 |         6 | Node 4 

 8 |         6 | Node 8 

11 |         6 | Node 11

 7 |        11 | Node 7 

10 |         7 | Node 10

db<>fiddle here