Creating an instance of a Binary Search Tree class when using generics?

Question

I have the following class :

public class BinarySearchTree<Key extends Comparable<? super Key>, E> 
{
private BTNode<Key, E> root;
int nodeCount;

/* Constructor */

public BinarySearchTree()
{
    this.root = null;
    this.nodeCount = 0;
}

...

I have no idea how to create an instance of it in my application though...

I have tried :

BinarySearchTree myTree = new BinarySearchTree();

and also,

BinarySearchTree<Integer> myTree = new BinarySearchTree<Integer>();

Any ideas are greatly welcomed!

where is BTNode defined?

George Simms
– George Simms

2015-07-24 19:13:26 +00:00
Commented Jul 24, 2015 at 19:13 — George Simms
– George Simms, Commented Jul 24, 2015 at 19:13

M A · Accepted Answer · 2015-07-24 18:06:16Z

1

Your BinarySearchTree has two type variables in it: one called Key for the comparable key, and one called E for the type of the node content. You're specifying just one type argument in the variable declaration:

BinarySearchTree<Integer, MyType> myTree = new BinarySearchTree<Integer, MyType>();

answered Jul 24, 2015 at 18:06

M A

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3 Comments

Slippy Over a year ago

That is exactly what I am looking for when creating the list! Thanks One other quick question though - Do you know how I might then call my insert? it is defined as public void insert(Key key, Element e)

M A Over a year ago

Something like myTree.insert(someInteger, myTypeInstance).

Slippy Over a year ago

Thanks! This helped a lot.

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