Tree structure in sql in Oracle.How to show tree,child nodes and parent nodes in SQL Oracle

Question

I would like to show a tree structure in SQL with child nodes and parent nodes. I have a table like:

Employee
-------------
ID (int)
FirstName (varchar)
LastName (varchar)
ParentID (int)
Job (varchar)

which represents an employee. ParentID represent the manager of the employee has. I would like to have this table only with this structure.

1. I would like to show the whole tree structure.
2. I would like to show only the children nodes
3. I would like to show only the parent nodes

SampleDataImage

Answer 1

Query - The whole tree structure:

SELECT *
FROM   Employee
START WITH ParentID IS NULL
CONNECT BY PRIOR ID = ParentID
ORDER SIBLINGS BY LastName, FirstName, ID;

Query - The children of a given employee:

You do not need a hierarchical query for this.
(The parent is given by the bind variable :parent_id)

SELECT *
FROM   Employee
WHERE  ParentID = :parent_id
ORDER BY LastName, FirstName, ID;

Query - The descendants of a given employee:

The same query as for the whole tree but with a different start point
(The parent is given by the bind variable :parent_id)

SELECT *
FROM   Employee
START WITH ParentID = :parent_id
CONNECT BY PRIOR ID = ParentID
ORDER SIBLINGS BY LastName, FirstName, ID;

Query - The employee and their ancestors:

Similar to the previous query but with the CONNECT BY reversed and you won't need to order the siblings as there will only be one immediate manager per employee.
(The employee is given by the bind variable :employee_id)

SELECT *
FROM   Employee
START WITH ID = :employee_id
CONNECT BY PRIOR ParentID = ID;

Query - The employee's manager:

Identical to the previous query but with a filter LEVEL = 2 to just get the immediate parent row.
(The employee is given by the bind variable :employee_id)

SELECT e.*
FROM   Employee e
WHERE  LEVEL = 2
START WITH ID = :employee_id
CONNECT BY PRIOR ParentID = ID;