CIS 2168 / 2: Lab-Assignment #6: Performance comparison of Arrays and Binary Search Trees

THIS IS A 1 WEEK ASSIGNMENT !

No graphics this time. Pure, simple, dry: data structure performance evaluation:

Task 1:

Build your own Binary Search Tree data structure class 'BST'. BST should contain:

field: NodeBST root; // the tree's root.
method: public void insert(double v); // inserts a value into the tree
method: public NodeBST find(double v); // finds a value in the tree and returns reference to node(or returns null)
method: public void traversePre(); // traverses the tree in pre-order and prints the value of each node

The node-class NodeBST you have to provide consists of:

field: double value; // the node's value
field: NodeBST ChildL; // left child
field: NodeBST ChildR; // right child

Task 2:

Insert the numbers 14,16,19,3,1,2,4,7 (in this order!) into the tree.
make a test print using your traversePre() function. It must show the numbers in the following order: 14,3,1,2,4,7,16,19. If it does not: go back to task 1 and fix your bug.
check if your find routine works: call "NodeBST n = find(7)" and see if n.value == 7. If it isn't: go back to task 1 and fix your bug.

Task 3:

Check how fast the tree performs:

for each number (14,3,..,19) call find() 10000 times and store the total time for all 80000 searches(time measure: see below).
since i didn't try this: if it's too slow, try 1000 times, if it's too fast, try 100000 times for each number. print the result for each number (e.g.: "10000 times searching for 14.0: xxx seconds")
now store the numbers in an array, and do the same search in the array, also storing the time.
compare the timing results (print total time for tree and array).

Task 4:

Check again how fast the tree performs, but this time the order of numbers is different:

erase the old tree, and create a new tree inserting the same numbers, but in a different order: 1,2,3,4,7,16,19
perform the timing comparison of Task 3.

How to measure time in JAVA

The method "static long System.currentTimeMillis()" returns the current time in milli seconds. To check how long a task needs, call that function before and after the task and store the time difference.

Enjoy!