Priority Queue not understanding how to trace the alogrithm

Question

These are the instructions that I received...

Insert the following n objects, in the order given, into a binary min-heap, you should trace the push method. 5, 3, 9, 7, 2, 4, 6, 1, 8
Apply the pop() method 3 times

i'm not understanding how pop from my binary tree

this is what I have..

This is the code that I need to trace and figure out how to pop.

import java.util.Collection;
import java.util.NoSuchElementException;

/**
 * PriorityQueue class implemented via the binary heap.
 */
public class PriorityQueue<AnyType>
{

    private static int INITSIZE = 100;

    private int currentSize;   // Number of elements in heap
    private AnyType [ ] array; // The heap array


    /**
     * Construct an empty PriorityQueue.
     */
    public PriorityQueue( )
    {
        currentSize = 0;
        array = (AnyType[]) new Object[ INITSIZE + 1 ];
    }


    /**
     * Compares lhs and rhs using compareTo
     */
    private int compare( AnyType lhs, AnyType rhs ) {
        return ((Comparable)lhs).compareTo( rhs );  
    }

    /**
     * Inserts an item to this PriorityQueue.
     * @param x any object.
     * @return true.
     */
    public boolean push( AnyType x )
    {
        if( currentSize + 1 == array.length )
            expandArray( );

            // Percolate up
        int hole = ++currentSize;
        array[ 0 ] = x;

        for( ; compare( x, array[ hole / 2 ] ) < 0; hole /= 2 )
            array[ hole ] = array[ hole / 2 ];
        array[ hole ] = x;

        return true;
    }

    /**
     *  isEmpty() indicates whether the heap is empty.
     *  @return true if the list is empty, false otherwise.
     **/

    public boolean isEmpty() {
        return currentSize == 0;
    }

    /**
     * Returns the number of items in this PriorityQueue.
     * @return the number of items in this PriorityQueue.
     */
    public int size( )
    {
        return currentSize;
    }

    /**
     * Make this PriorityQueue empty.
     */
    public void clear( )
    {
        currentSize = 0;
    }

    /**
     * Returns the smallest item in the priority queue.
     * @return the smallest item.
     * @throws NoSuchElementException if empty.
     */
    public AnyType element( )
    {
        if( isEmpty( ) )
            throw new NoSuchElementException( );
        return array[ 1 ];
    }

    /**
     * Removes the smallest item in the priority queue.
     * @return the smallest item.
     * @throws NoSuchElementException if empty.
     */
    public AnyType pop( )
    {
        AnyType minItem = element( );
        array[ 1 ] = array[ currentSize-- ];
        percolateDown( 1 );

        return minItem;
    }


    /**
     * Establish heap order property from an arbitrary
     * arrangement of items. Runs in linear time.
     */
    private void buildHeap( )
    {
        for( int i = currentSize / 2; i > 0; i-- )
            percolateDown( i );
    }


    /**
     * Internal method to percolate down in the heap.
     * @param hole the index at which the percolate begins.
     */
    private void percolateDown( int hole )
    {
        int child;
        AnyType tmp = array[ hole ];

        for( ; hole * 2 <= currentSize; hole = child )
        {
            child = hole * 2;
            if( child != currentSize &&
                    compare( array[ child + 1 ], array[ child ] ) < 0 )
                child++;
            if( compare( array[ child ], tmp ) < 0 )
                array[ hole ] = array[ child ];
            else
                break;
        }
        array[ hole ] = tmp;
    }

    /**
     *  expandArray(): internal method to extend array.
     *  creates a new array with larger size (twice)
     */
    private void expandArray() {
        AnyType [ ] newArray;

        newArray = (AnyType []) new Object[ array.length * 2 ];
        for( int i = 0; i < array.length; i++ )
            newArray[ i ] = array[ i ];
        array = newArray;
    }

    public static void main( String [ ] args )
    {
        PriorityQueue t = new PriorityQueue( );
        final int NUMS = 4000;
        final int GAP  =   37;

        System.out.println( "Checking... (no more output means success)" );
        int min = 1000000;
        for( int i = GAP; i != 0; i = ( i + GAP ) % NUMS ) {
            if (min > i)
                min = i;
            t.push( new Integer( i ) );
            if( ((Integer)(t.element( ))).intValue( ) != min )
                System.out.println( "Push error! "+i+"   "
                        +((Integer)(t.element( ))).intValue( ));
        }

        for( int i = 1; i < NUMS; i++ )
             if( ((Integer)(t.pop( ))).intValue( ) != i )
                 System.out.println( "Pop error!" );
    }
}

What have you tried so far, and what exactly was the problem? — Mureinik, Mar 23 '18 at 20:35
What did you do until now? Please show your code as a [MCVE](https://stackoverflow.com/help/mcve) — milbrandt, Mar 23 '18 at 20:41
I don't understand how to trace the numbers through the program. — Issajatt, Mar 23 '18 at 20:54
System.out for stuff you don't know. Otherwise, probably task assignment ask you to show tree after first 3 pops... — Andrej Buday, Mar 23 '18 at 22:18
If this is supposed to be a min-heap, why are you drawing it with 9 at the root? See https://www.cs.usfca.edu/~galles/JavascriptVisual/Heap.html — pjs, Mar 23 '18 at 22:29

Jim Mischel · Accepted Answer · 2018-03-26T02:09:37.617

The procedure for removing the first item from a heap is:

1. Copy the first item from the array. That is the return value.
2. Take the lowest, right-most node (in this case, it would be the value 3, and place it in the first position of the array.
3. Reduce the number of items in the array by 1.
4. Sift the item down from the root to its proper place.

By the way, what you have there is a max heap. In a min heap, the smallest item is at the root.

See my blog post, A better way to do it: the heap, for a more detailed discussion of how to manipulate a binary heap.

Priority Queue not understanding how to trace the alogrithm

1 Answers1