How many elements can be held in a B-tree of order n?

Question

Is it 2n? Just checking.

Answer 1

Terminology
The Order of a B-Tree is inconstantly defined in the literature.
(see for example the terminology section of Wikipedia's article on B-Trees)
Some authors consider it to be the minimum number of keys a non-leaf node may hold, while others consider it to be the maximum number of children nodes a non-leaf node may hold (which is one more than the maximum number of keys such a node could hold).
Yet many others skirt around the ambiguity by assuming a fixed length key (and fixed sized nodes), which makes the minimum and maximum the same, hence the two definitions of the order produce values that differ by 1 (as said the number of keys is always one less than the number of children.)

I define depth as the number of nodes found in the search path to a leaf record, and inclusive of the root node and the leaf node. In that sense, a very shallow tree with only a root node pointing directly to leaf nodes has depth 2. If that tree were to grow and require an intermediate level of non-leaf nodes, its depth would be 3 etc.

How many elements can be held in a B-Tree of order n?
Assuming fixed length keys, and assuming that "order" n is defined as the maximum number of child nodes, the answer is:

   (Average Number of elements that fit in one Leaf-node) * n ^ (depth - 1)

How do I figure?...:
The data (the "elements") is only held in leaf nodes. So the number of element held is the average number of elements that fit in one node, times the number of leaf nodes.
The number of leaf nodes is itself driven by the number of children that fit in a non-leaf node (the order). For example the non-leaf node just above a leaf node, points to n (the order) leaf-nodes. Then, the non-leaf node above this non-leaf node points to n similar nodes etc, hence "to the power of (depth -1)".

Note that the formula above generally holds using the averages (of key held in a non-leaf node, and of elements held in a leaf node) rather than assuming fixed key length and fixed record length: trees will typically have a node size that is commensurate with the key and record sizes, hence holding a number key or records that is big enough that the effective number of keys or record held in any leaf will vary relatively little compared with the average.

Example:
A tree of depth 4 (a root node, two level of non-leaf nodes and one level [obviously] of leaf nodes) and of order 12 (non-leaf nodes can hold up to 11 keys, hence point to 12 nodes below them) and such that leaf nodes can contain 5 element each, would:
  - have its root node point to 12 nodes below it     - each node below it points to 12 nodes below them (hence there will be 12 * 12 nodes in the layer "3" (assuming the root is layer 1 etc., this numbering btw is also ambiguously defined...)     - each node in "layer 3" will point to 12 leaf-nodes (hence there will be 12 * 12 * 12 leaf nodes.
  - each leaf node has 5 elements (in this example case)
Hence.. such a tree will hold...

  Nb Of Elements in said tree = 5 * 12 * 12 * 12
                              = 5 * (12 ^ 3)
                              = 5 * (12 ^ depth -1)
                              = 8640

Recognize the formula on the 3rd line.

What is generally remarkable for B-Tree, and which makes for their popularity is that a relatively shallow tree (one with a limited number of "hops" between the root and the sought record), can hold a relatively high number record. This number is multiplied by the order at each level.