computer notes - data structures - 26

8/3/2019 Computer Notes - Data Structures - 26

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Class No.26

Data Structures

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Heap code in C++

template <class eType>

void Heap<eType>::deleteMin( eType & minItem )

{

if( isEmpty( ) ) {

cout << "heap is empty.“ << endl

;return;

}

minItem = array[ 1 ];

array[ 1 ] = array[ currentSize-- ];

percolateDown( 1 );

}






















Heap code in C++

// hole is the index at which the percolate begins.

template <class eType> void Heap<eType>::percolateDown( int hole )

{

int child;

eType tmp = array[ hole ];

for( ; hole * 2 <= currentSize; hole = child )

{child = hole * 2;

if( child != currentSize &&

array[child+1] < array[ child ] )

child++; // right child is smaller

if( array[ child ] < tmp )

array[ hole ] = array[ child ];

else break;

}

array[ hole ] = tmp;

}






Heap code in C++


const eType& Heap<eType>::getMin( )

{

if( !isEmpty( ) )

return array[ 1 ];

}


void Heap<eType>::buildHeap(eType* anArray, int n )

{

for(int i = 1; i <= n; i++)

array[i] = anArray[i-1];currentSize = n;

for( int i = currentSize / 2; i > 0; i-- )

percolateDown( i );

}






Heap code in C++


bool Heap<eType>::isEmpty( )

{

return currentSize == 0;

}


bool Heap<eType>::isFull( ){

return currentSize == capacity;

}


int Heap<eType>::getSize( ){

return currentSize;

}






BuildHeap in Linear Time

How is buildHeap a linear time algorithm? I.e.,better than Nlog2N?

We need to show that the sum of heights is alinear function of N (number of nodes).

Theorem :For a perfect binary tree of height h containing

2h +1

–

1 nodes, the sum of the heights of nodesis 2h +1 – 1 – (h +1), or N -h-1.







It is easy to see that this tree consists of (20)node at height h , 21 nodes at height h – 1, 22

at h- 2 and, in general, 2i

nodes at h – i.






Complete Binary Tree

A

B

h : 20 nodes

H

D

I

E

J K

C

L

F

M

G

N O

h -1: 21

nodes

h -2: 22 nodes

h -3: 23 nodes







The sum of the heights of all the nodes is then

S = 2i (h – i ), for i = 0 to h-1

= h + 2 (h- 1) + 4 (h- 2 ) + 8 (h- 3 )+ ….. + 2 h-1 (1)

Multiplying by 2 gives the equation

2S = 2h + 4 (h- 1) + 8 (h- 2 ) + 16 (h- 3 )+ ….. + 2 h (2)

Subtract the two equations to get

S = -h + 2 + 4 + 8 + 16 + ….. + 2 h-1 + 2 h

= (2 h+1 – 1) - (h+ 1)

Which proves the theorem.








Since a complete binary tree has between 2 h

and 2 h+1 nodes

S = (2 h+1

– 1) - (h+ 1) N - log 2 ( N +1)

Clearly, as N gets larger, the log2(N +1) term

becomes insignificant and S becomes afunction of N .







Another way to prove the theorem.

The height of a node in the tree = the number ofedges on the longest downward path to a leaf

The height of a tree = the height of its root

For any node in the tree that has some height h ,darken h tree edges

– Go down tree by traversing left edge then only

right edges There are N – 1 tree edges, and h edges on right

path, so number of darkened edges is N – 1 – h ,which proves the theorem.







Height 1 Nodes

Marking the left edges for height 1 nodes http://ecomputernotes.com





Height 2 Nodes

Marking the first left edge and the subsequent right edge for height 2

nodes






Height 3 Nodes

Marking the first left edge and the subsequent two right edges for

height 3 nodes







Height 4 Nodes

Marking the first left edge and the subsequent three right edges for

height 4 nodes







Theorem

N=31, treeEdges=30, H=4, dottedEdges=4 (H).

Darkened Edges = 26 = N-H-1 (31-4-1)




computer notes - data structures - 26

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