There are many ways to represent a tree or hierarchy in SQL. This is
called an adjacency list model and it looks like this:
CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
boss CHAR(10) DEFAULT NULL REFERENCES OrgChart(emp),
salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);
OrgChart
emp boss salary
===========================
'Albert' 'NULL' 1000.00
'Bert' 'Albert' 900.00
'Chuck' 'Albert' 900.00
'Donna' 'Chuck' 800.00
'Eddie' 'Chuck' 700.00
'Fred' 'Chuck' 600.00
Another way of representing trees is to show them as nested sets.
Since SQL is a set oriented language, this is a better model than the
usual adjacency list approach you see in most text books. Let us define
a simple OrgChart table like this.
CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt) );
OrgChart
emp lft rgt
======================
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10
The organizational chart would look like this as a directed graph:
Albert (1, 12)
/ \
/ \
Bert (2, 3) Chuck (4, 11)
/ | \
/ | \
/ | \
/ | \
Donna (5, 6) Eddie (7, 8) Fred (9, 10)
The adjacency list table is denormalized in several ways. We are
modeling both the Personnel and the organizational chart in one table.
But for the sake of saving space, pretend that the names are job titles
and that we have another table which describes the Personnel that hold
those positions.
Another problem with the adjacency list model is that the boss and
employee columns are the same kind of thing (i.e. names of personnel),
and therefore should be shown in only one column in a normalized table.
To prove that this is not normalized, assume that "Chuck" changes his
name to "Charles"; you have to change his name in both columns and
several places. The defining characteristic of a normalized table is
that you have one fact, one place, one time.
The final problem is that the adjacency list model does not model
subordination. Authority flows downhill in a hierarchy, but If I fire
Chuck, I disconnect all of his subordinates from Albert. There are
situations (i.e. water pipes) where this is true, but that is not the
expected situation in this case.
To show a tree as nested sets, replace the nodes with ovals, and then
nest subordinate ovals inside each other. The root will be the largest
oval and will contain every other node. The leaf nodes will be the
innermost ovals with nothing else inside them and the nesting will show
the hierarchical relationship. The (lft, rgt) columns (I cannot use the
reserved words LEFT and RIGHT in SQL) are what show the nesting. This is
like XML, HTML or parentheses.
At this point, the boss column is both redundant and denormalized, so it
can be dropped. Also, note that the tree structure can be kept in one
table and all the information about a node can be put in a second table
and they can be joined on employee number for queries.
To convert the graph into a nested sets model think of a little worm
crawling along the tree. The worm starts at the top, the root, makes a
complete trip around the tree. When he comes to a node, he puts a number
in the cell on the side that he is visiting and increments his counter.
Each node will get two numbers, one of the right side and one for the
left. Computer Science majors will recognize this as a modified preorder
tree traversal algorithm. Finally, drop the unneeded OrgChart.boss
column which used to represent the edges of a graph.
This has some predictable results that we can use for building queries.
The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM
TreeTable)); leaf nodes always have (left + 1 = right); subtrees are
defined by the BETWEEN predicate; etc. Here are two common queries which
can be used to build others:
1. An employee and all their Supervisors, no matter how deep the tree.
SELECT O2.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp = :myemployee;
2. The employee and all subordinates. There is a nice symmetry here.
SELECT O1.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O2.emp = :myemployee;
3. Add a GROUP BY and aggregate functions to these basic queries and you
have hierarchical reports. For example, the total salaries which each
employee controls:
SELECT O2.emp, SUM(S1.salary)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp = S1.emp
GROUP BY O2.emp;
4. To find the level of each emp, so you can print the tree as an
indented listing. Technically, you should declare a cursor to go with
the ORDER BY clause.
SELECT COUNT(O2.emp) AS indentation, O1.emp
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
GROUP BY O1.lft, O1.emp
ORDER BY O1.lft;
5. The nested set model has an implied ordering of siblings which the
adjacency list model does not. To insert a new node, G1, under part G.
We can insert one node at a time like this:
BEGIN ATOMIC
DECLARE rightmost_spread INTEGER;
SET rightmost_spread
= (SELECT rgt
FROM Frammis
WHERE part = 'G');
UPDATE Frammis
SET lft = CASE WHEN lft > rightmost_spread
THEN lft + 2
ELSE lft END,
rgt = CASE WHEN rgt >= rightmost_spread
THEN rgt + 2
ELSE rgt END
WHERE rgt >= rightmost_spread;
INSERT INTO Frammis (part, lft, rgt)
VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
COMMIT WORK;
END;
The idea is to spread the (lft, rgt) numbers after the youngest child of
the parent, G in this case, over by two to make room for the new
addition, G1. This procedure will add the new node to the rightmost
child position, which helps to preserve the idea of an age order among
the siblings.
6. To convert a nested sets model into an adjacency list model:
SELECT B.emp AS boss, E.emp
FROM OrgChart AS E
LEFT OUTER JOIN
OrgChart AS B
ON B.lft
= (SELECT MAX(lft)
FROM OrgChart AS S
WHERE E.lft > S.lft
AND E.lft < S.rgt);
7. To convert an adjacency list to a nested set model, use a push down
stack. Here is version with a stack in SQL/PSM.
-- Tree holds the adjacency model
CREATE TABLE Tree
(node CHAR(10) NOT NULL,
parent CHAR(10));
-- Stack starts empty, will holds the nested set model
CREATE TABLE Stack
(stack_top INTEGER NOT NULL,
node CHAR(10) NOT NULL,
lft INTEGER,
rgt INTEGER);
CREATE PROCEDURE TreeTraversal ()
LANGUAGE SQL
DETERMINISTIC
BEGIN ATOMIC
DECLARE counter INTEGER;
DECLARE max_counter INTEGER;
DECLARE current_top INTEGER;
SET counter = 2;
SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
SET current_top = 1;
--clear the stack
DELETE FROM Stack;
-- push the root
INSERT INTO Stack
SELECT 1, node, 1, max_counter
FROM Tree
WHERE parent IS NULL;
-- delete rows from tree as they are used
DELETE FROM Tree WHERE parent IS NULL;
WHILE counter <= max_counter- 1
DO IF EXISTS (SELECT *
FROM Stack AS S1, Tree AS T1
WHERE S1.node = T1.parent
AND S1.stack_top = current_top)
THEN BEGIN -- push when top has subordinates and set lft value
INSERT INTO Stack
SELECT (current_top + 1), MIN(T1.node), counter, NULL
FROM Stack AS S1, Tree AS T1
WHERE S1.node = T1.parent
AND S1.stack_top = current_top;
-- delete rows from tree as they are used
DELETE FROM Tree
WHERE node = (SELECT node
FROM Stack
WHERE stack_top = current_top + 1);
-- housekeeping of stack pointers and counter
SET counter = counter + 1;
SET current_top = current_top + 1;
END;
ELSE
BEGIN -- pop the stack and set rgt value
UPDATE Stack
SET rgt = counter,
stack_top = -stack_top -- pops the stack
WHERE stack_top = current_top;
SET counter = counter + 1;
SET current_top = current_top - 1;
END;
END IF;
-- the top column is not needed in the final answer
END WHILE;
-- SELECT node, lft, rgt FROM Stack;
END;
I have a book on TREES & HIERARCHIES IN SQL coming out in April 2004.
--CELKO--
===========================
Please post DDL, so that people do not have to guess what the keys,
constraints, Declarative Referential Integrity, datatypes, etc. in your
schema are.
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