CSCI-1200 Homework 8 Simplified B+ Trees solved


Original Work ?


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In this assignment we will be implementing a partial and modified version of B+ trees. As a result, online
resources may not use the same terminology or may describe implementation details that are not relevant
to our HW8 implementation.

You should read the entire assignment before beginning your work. You
should also make sure you understand the basic concepts discussed at the end of Lecture 17. It is highly
recommended that before you begin coding, you practice constructing a couple of trees (using b = 3, b = 4)
by hand and then checking your work with this online visualization tool:

The bulk of the assignment will focus on proper insertion in a B+ tree, which is described on the next page.

Implementation Details

In this assignment we will assume that the keys we insert are unique, i.e. for a particular B+ tree, we will
never call insert(3); insert(3);. We will also assume that b > 2 in all tests. You will find it beneficial to
borrow code from our partial implementation of the ds set class.

We will not implement iterators, so find()
should instead return a pointer to a BPlusTreeNode. If the tree is empty, this will be a NULL pointer,
otherwise this will be the leaf node where the key is/would be.

The print functions only need to work with
types/classes that already work with operator<<, and PrintSideways makes its split at b/2 children. You
must implement all of the functions required to make hw8 test.cpp compile and run correctly.


Unless the tree is empty, find() will always return a pointer to a node in the tree. You do not need to store
NULL pointers. In the middle of an insertion, it is okay to let your nodes hold too many keys or children as
long as you fix it before the insertion and splits are finished. Since this is a tree, some things will be more
“natural” to do with recursion.


While you are encouraged to write your own test functions, we will not be looking at them. You only need
to submit a README.txt and BPlusTree.h file. Dr. Memory will be used on this assignment and you will
be penalized for leaks and memory errors in your solution. If you get at least 6 points between test cases
4, 5, and 6 by the end of Wednesday, you can submit on Friday without being charged a late day. Please
remember that all submissions are still due by the end of Saturday.

Extra Credit

With the way print BFS() is currently expected to output, it is not possible to tell which nodes are children
of a particular node. Assuming that each key is short (i.e. no more than 2 characters wide), implement a
function, print BFS pretty() that still uses a BFS ordering and a vertical layout like print BFS(), but that
has appropriate spacing so the structure of the tree is apparent.

There are several possible ways to handle
this, so you may choose whatever design you think is reasonable. Make sure to leave a note in your README
if you implement the extra credit.

a b c
a b
a b c

Starting from an empty tree, with b = 3 in this example we do
the following:
(1) insert(b); creates a root node with “b” in it.
(2) insert(a); adds “a” to the root node
(3) insert(c); adds “c” to the root node, which makes it too
(4) The root node splits into two nodes, one with “a” and one
with “b”, and “c”. A new parent is created, with the first
value from the new right-hand node (“b”) placed in it. A
node split should create two new nodes and put half of the
orginal node’s keys into each of the new nodes. Whenever
there are an odd number of keys in a node that needs to be
split, the “extra” key will go to the right-hand node.

(1) This example starts with one possible
tree with b = 3 and the keys “a”-“f”.
(2) insert(ant) causes a leaf to become
too full. “ant” comes after “a”
according to string’s operator<.
(3) The leaf containing “a”, “ant”, “b” is
split into two nodes, but this makes
the parent too full.

(4) We split the overfull node, creating
a new parent which is now the root.
Since this split a non-leaf node, we do
not copy the middle key of “a,c,e” into
the new left/right nodes – “c” only
appears in the newly created root. A
split at a leaf always has the potential
to cause more splits all the way up to
splitting the root.
a b
c e
c d e f
c e
a ant b c d e f
a ant b c d e f
ant e
a ant b c d e f
ant c e