Re: How to express a b-tree in SQL?

>> I was wondering if anybody has an opinons on the best way to express
a binary tree data structure in a relational table. Also if anybody could point me to a book or online document which discusses how to do this I would be very interested. <<

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 Personnel table like this, ignoring the left (lft) and right (rgt) columns for now.

This problem is always given with a column for the employee and one for his boss in the textbooks:

 CREATE TABLE Personnel
 (emp CHAR(10) PRIMARY KEY,
  boss CHAR(10), -- this column is unneeded & denormalizes the table
  salary DECIMAL(6,2) NOT NULL,
  lft INTEGER NOT NULL,
  rgt INTEGER NOT NULL);

 Personnel
 emp boss salary lft rgt

 Albert   NULL     1000.00   1   12
 Bert     Albert    900.00   2    3
 Chuck    Albert    900.00   4   11
 Donna    Chuck     800.00   5    6
 Eddie    Chuck     700.00   7    8
 Fred     Chuck     600.00   9   10

 which 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)

This (without the lft and rgt columns) is called the adjacency list model, after the graph theory technique of the same name; the pairs of nodes are adjacent to each other. The 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.

To show a tree as nested sets, replace the nodes with ovals, 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 rgt and lft columns (I cannot use the reserved words LEFT and RIGHT in SQL) are what shows the nesting.

If that mental model does not work, then imagine a little worm crawling anti-clockwise along the tree. Every time he gets to the left or right side of a node, he numbers it. The worm stops when he gets all the way around the tree and back to the top.

This is a natural way to model a parts explosion, since a final assembly is made of physically nested assemblies that final break down into separate parts.

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 Personnel.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 P2.*
   FROM Personnel AS P1, Personnel AS P2   WHERE P1.lft BETWEEN P2.lft AND P2.rgt     AND P1.emp = :myemployee;

2. The employee and all subordinates. There is a nice symmetry here.

 SELECT P2.*
   FROM Personnel AS P1, Personnel AS P2   WHERE P1.lft BETWEEN P2.lft AND P2.rgt     AND P2.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 P2.emp, SUM(P1.salary)
   FROM Personnel AS P1, Personnel AS P2   WHERE P1.lft BETWEEN P2.lft AND P2.rgt   GROUP BY P2.emp;

4. Finding all leaf nodes is trivial:

 SELECT P1.*
   FROM Personnel AS P1
  WHERE rgt = lft - 1;

This will be two to three orders of magnitude faster than the adjacency list model.

For details, see the chapter in my book SQL FOR SMARTIES (second editon) by Joe Celko, Morgan-Kaufmann, 1995, ISBN 1-55860-323-9. For a binary tree, you will need to add some constraints about the relationship between lft and rgt values of siblings. Play with it and let me know if you can come up with a good constraint and Iw ill use it in the next edition of the book.

-CELKO-- Sent via Deja.com http://www.deja.com/
Before you buy. Received on Tue Feb 29 2000 - 00:00:00 CET