Re: BinaryTree

Jeff Higgins wrote:


<http://en.wikipedia.org/wiki/Binary_tree>
<http://en.wikipedia.org/wiki/Binary_search_tree>

Within the tree, each node has a value and a link set, since you say you want
to be link-based. I don't really think "linked list" applies; you are talking
about a "linked tree".

Here's an example implementation that permits only one node with a given value
in a tree:

<sscce>
package testit;
public class Node <T extends Comparable <? super T>>
    implements Comparable <Node <T>>
{
   private final T value;
   private Node <T> left, right; // the links

   public Node( final T v )
   {
     this.value = v;
   }

   public final T getValue()
   { return value; }

   public final Node <T> getLeft()
   { return left; }
   public final void setLeft( Node <T> v )
   { left = v; }

   public final Node <T> getRight()
   { return right; }
   public final void setRight( Node <T> v )
   { right = v; }

   public int compareTo( Node <T> o )
   {
     return (this == o? 0
         : o == null? 1
         : value == null? (o.value == null? 0 : -1)
         : o.value == null? 1 /* guarantee symmetry */
         : value.compareTo( o.value ));
   }

   @Override public boolean equals( Object o )
   {
     try
     {
       return equals( getClass().cast(o) );
     }
     catch ( ClassCastException exc )
     {
       return false;
     }
   }

   public boolean equals( Node <T> o )
   {
     return (this == o || (o != null
        && (value == null? o.value == null : value.equals( o.value ))
        ));
   }

   @Override public int hashCode()
   {
     return (value == null? 0 : value.hashCode());
   }

   @Override public String toString()
   {
     return value.toString();
   }
}
</sscce>

--
Lew