What are the differences between B trees and B+ trees

Assuming you're joining on columns with no duplicates, which is a very common case:

• An inner join of A and B gives the result of A intersect B, i.e. the inner part of a Venn diagram intersection.

• An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.

Examples

Suppose you have two tables, with a single column each, and data as follows:

A    B
-    -
1    3
2    4
3    5
4    6

Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.

Inner join

An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.

select * from a INNER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b;

a | b
--+--
3 | 3
4 | 4

Left outer join

A left outer join will give all rows in A, plus any common rows in B.

select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b(+);

a |  b
--+-----
1 | null
2 | null
3 |    3
4 |    4

Right outer join

A right outer join will give all rows in B, plus any common rows in A.

select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a(+) = b.b;

a    |  b
-----+----
3    |  3
4    |  4
null |  5
null |  6

Full outer join

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.

select * from a FULL OUTER JOIN b on a.a = b.b;

 a   |  b
-----+-----
   1 | null
   2 | null
   3 |    3
   4 |    4
null |    6
null |    5

My best advice would be that there is no standard tree data structure because there are so many ways you could implement it that it would be impossible to cover all bases with one solution. The more specific a solution, the less likely it is applicable to any given problem. I even get annoyed with LinkedList - what if I want a circular linked list?

The basic structure you'll need to implement will be a collection of nodes, and here are some options to get you started. Let's assume that the class Node is the base class of the entire solution.

If you need to only navigate down the tree, then a Node class needs a List of children.

If you need to navigate up the tree, then the Node class needs a link to its parent node.

Build an AddChild method that takes care of all the minutia of these two points and any other business logic that must be implemented (child limits, sorting the children, etc.)