InOrderIterator.java

/*
  Copyright (c) 1999, 2000 Brown University, Providence, RI
  
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  documentation for any purpose other than its incorporation into a
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  copyright notice and this permission notice appear in supporting
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  without specific, written prior permission.
  
  BROWN UNIVERSITY DISCLAIMS ALL WARRANTIES WITH REGARD TO THIS
  SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND
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  UNIVERSITY BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL
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  PERFORMANCE OF THIS SOFTWARE.
*/

package jdsl.core.ref;

import jdsl.core.api.*;
import java.util.NoSuchElementException;



/** 
 * The inorder iterator gives an inorder iteration of a binary tree.
 *
 * Creating this iterator takes O(N) where N = the number of positions in the tree,
 * assuming that root, rightChild, leftChild, isInternal are O(1) in the tree
 * implementation.
 * All other methods take O(1) time
 *
 * @author Ryan Shaun Baker
 * @author Mike Boilen (mgb)
 * @version JDSL 2.1.1 
 */
public class InOrderIterator implements PositionIterator{
  /**
    * The array of positions in the tree
    */
  private Position[] positions_;

  /**
    * The tree we are iterating over
    * (only used during construction of array)
    */
  private InspectableBinaryTree tree_;

  /**
    * The current Index
    */
  private int iCurrentIndex_;

  /**
    * The last index of the array
    */
  private int iLastIndex_;

    /** 
     *Constructs a new InOrderIterator to iterate over the given container
     * Puts a reference to each position into the array --
     * takes O(N) time where N = the number of positions in the container
     * (given assumptions discussed in header)
     * @param tree The tree to iterate over
     */
    public InOrderIterator(InspectableBinaryTree tree) {
        tree_ = tree;
	positions_= new Position[tree.size()];
	iCurrentIndex_=0;
	iLastIndex_=positions_.length-1;	

	traverse(tree.root());
	iCurrentIndex_=-1;
    }
    
  /**
     * Traverses the tree to fill the array
     * takes O(N) time when called on the root
     * where N = the number of positions in the container
     * and O(S) time when called on another position
     * where S = the number of positions in the subtree rooted at that position
     *
     * @param curpos Our current position along the tree
     */
  private void traverse(Position curpos){
    boolean isInt = false;
    if (tree_.isInternal(curpos)){
      isInt= true;
      traverse(tree_.leftChild(curpos));
    }
    positions_[iCurrentIndex_++] = curpos;
    if (isInt)
      traverse(tree_.rightChild(curpos));
  }

  /**
    * Takes O(1) time
    */
  public boolean hasNext()
    {
      return iCurrentIndex_<iLastIndex_;
    }

  /**
    * Takes O(1) time
    */
    public Object nextObject()
    {
      if(!hasNext()) {
	throw new NoSuchElementException("End of iterator contents reached");
      }
      return (positions_[++iCurrentIndex_]);
    }

 /**
   * Takes O(1) time
   */
  public Position nextPosition () {
    if(!hasNext()) {
      throw new NoSuchElementException("End of iterator contents reached");
    }
    return (positions_[++iCurrentIndex_]);
  }

  /**
    * Takes O(1) time
    */
  public Object object () {
    return position();
  }

  /**
    * Takes O(1) time
    */
  public Object element() throws NoSuchElementException{
    checkPastEnd();
    return position().element();
  }


  /**
    * Takes O(1) time
    */
  public Position position() throws NoSuchElementException{    
    checkPastEnd();
    return positions_[iCurrentIndex_];
  }


  /** 
   * Takes O(1) time
   * Sets the current node to the first node.
   */
  public void reset (){
    iCurrentIndex_ = -1;
  }


  /**
    * Takes O(1) time
    * Checks if we're past the end of the iterator
    */
  private void checkPastEnd() throws NoSuchElementException{
    if (iCurrentIndex_>iLastIndex_)
      throw new NoSuchElementException("Past End");
    if (iCurrentIndex_<0)
      throw new NoSuchElementException("You need to call next() at least once before you can call element() or position().");
  }
    
    


}