Re: help with nested set
Date: Tue, 14 Apr 2009 06:18:59 -0700 (PDT)
Message-ID: <274ae33c-39ad-441e-8f6c-ca0787abf0ad_at_k8g2000yqn.googlegroups.com>
>> For any client, a unit might belong under various organizational levels. ..I'm going to assume for now that they're all strictly hierarchical -- no overlapping. And, there may be any number of virtual grouping levels... I've been looking at both adjacency list and the preorder tree traversal algorithm <<
Here is a skeleton for the Nested Sets model
CREATE TABLE ClientOrgCharts
(client_id CHAR(9) NOT NULL UNIQUE
REFERENCES Client(client_id),
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt));
>> All child nodes *should* have the same client_id as the ultimate root, but I can't think of any way to ensure this in the DDL. I'll just have to "be careful" in code. <<
Look at the Nested Sets skeleton. No cycles since each client_id appears once. In a full SQL you can also add more constriants to assure a proper tree strcuture. Or use a VIEW and a WITH CHECK OPTION. The adjacency list reqiures cursors to do data integrity. Here is my "cut & paste" on tress:
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_name CHAR(10) NOT NULL PRIMARY KEY,
boss_emp_name CHAR(10) REFERENCES OrgChart(emp_name),
salary_amt DECIMAL(6,2) DEFAULT 100.00 NOT NULL,
<< horrible cycle constaints >>);
OrgChart
emp_name boss_emp_name salary_amt
'Albert' NULL 1000.00
'Bert' 'Albert' 900.00
'Chuck' 'Albert' 900.00
'Donna' 'Chuck' 800.00
'Eddie' 'Chuck' 700.00
'Fred' 'Chuck' 600.00
This approach will wind up with really ugly code -- CTEs hiding recursive procedures, horrible cycle prevention code, etc. The root of your problem is not knowing that rows are not records, that SQL uses sets and trying to fake pointer chains with some vague, magical non-relational "id".
This matches the way we did it in old file systems with pointer chains. Non-RDBMS programmers are comfortable with it because it looks familiar -- it looks like records and not rows.
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_name 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_name lft rgt
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10
The (lft, rgt) pairs are like tags in a mark-up language, or parens in algebra, BEGIN-END blocks in Algol-family programming languages, etc. -- they bracket a sub-set. This is a set-oriented approach to trees in a set-oriented language.
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_emp_name and employee columns are the same kind of thing (i.e. identifiers 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_emp_name 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_emp_name 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:
- 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_name = :myemployee;
2. The employee and all their 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_name = :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_name, SUM(S1.salary_amt)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp_name = S1.emp_name
GROUP BY O2.emp_name;
4. To find the level of each emp_name, so you can print the tree as an indented listing.
SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END) AS lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.lft
GROUP BY T1.node;
An untested version of this using OLAP functions might be better able to use the ordering.
SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END)
OVER (ORDER BY T1.lft
RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS
lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.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_name AS boss_emp_name, E.emp_name 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 find the immediate parent of a node:
SELECT MAX(P2.lft), MIN(P2.rgt)
FROM Personnel AS P!, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp_name = :my_emp_name;
8. 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 O1.
WHERE S1.node = O1.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(O1.node), counter, NULL
FROM Stack AS S1, Tree AS O1.
WHERE S1.node = O1.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;
END WHILE;
-- SELECT node, lft, rgt FROM Stack; -- the top column is not needed in the final answer -- move stack contents to new tree tableEND; I have a book on TREES & HIERARCHIES IN SQL which you can get at Amazon.com right now.
For a good article on using CTEs and recursion, see: http://www.sqlservercentral.com/columnists/fBROUARD/recursivequeriesinsql1999andsqlserver2005.asp Received on Tue Apr 14 2009 - 15:18:59 CEST