6 8 9 11 13 11 Heap: A Special Kind of Tree 15 26 16 25.

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Heap: A Special Kind of Tree

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Heap Order PropertyAt every node X, values inboth subtrees (if any)are greater than or equal tothe value in X.

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Heap Structure PropertyA heap is also a complete tree. “Complete” means completely filled, except perhaps for the bottom row, which is filled from left to right.


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Heap Structure PropertyA heap is also a completetree. “Complete” means completely filled, except perhaps for the bottom row, which is filled from left to right.

This is a binary heap.


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This is not a binary heap.

Why not?


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Heap orderviolation

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Ok

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Why not?


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Bottom rownot filledleft-to-right

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Why not?

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This is not a binary heap.Interior

row notfilled.

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This is a binary heap.

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Binary Search Tree PropertyAt every node X, values inleft subtree are smaller thanthe value in X, and values in right subtree are largerthan the value in X.


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versus

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versus

This is a heap.

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versus

This is a heap. It is not abinary searchtree.




versus

This is nota heap.

It is abinary searchtree.

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Leftist Heaps: Easily Mergeable Trees

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Leftist Structure PropertyFor every node X, the nullpath length of the left childis at least as large as that of the right child.



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The null path length,npl(X), of any node X isthe length of the shortestpath from X to a node withzero or one child.



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npl(X) = min(npl(left(X)), npl(right(X))) + 1npl() = 1



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Leftist Structure Property

npl(left(X)) npl(right(X))


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0 0

00 0

1 0

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2This is a leftist heap.


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0 0

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1 0

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2This is a leftist heap.

Note that itis also abinary heap.


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0 0

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1 0

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2This is a leftist heap.Suppose we

remove thisnode Note that it

is also abinary heap.


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0 0

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0 0

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Is the tree stillleftist?


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Yes.


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Is it stilla binaryheap?


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No,because. . .


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No,becauseit is notcomplete.


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Complete trees have good balance,but . . .


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Complete trees have good balance,but leftist trees can be very unbalanced.


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Although leftist heaps can be very unbalanced, they are easy to merge.

As a trivial case,consider two entirelyleft-leaning leftist heaps:


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To merge them,. . . 7

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To merge them,first make the onewith the largerroot node valuethe rightchildof theother.


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In the case shown,the result is alreadya leftist heap,and the mergeis finished.


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60If we had started with

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. . . then after the firststep the tree is not leftist, since the left child of the root hasnull path length–1, which is less thanthat of theright subtree.

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In this case, the mergealgorithm swapsthe left and rightsubtrees of theroot.



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In either case, the mergeis done inconstant time!



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Remember, though, thesewere special cases(entirely left-leaningtrees).


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In general, the mergecan’t be done in constant time,but . . .



In general, the mergecan’t be done in constant time,but it can be donein O(log N) time,where N isthe number ofnodes in thelarger of the twoheaps.

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T (H , H ) =

O(max(r , r ))


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r = rpl(H ) + 1ii

merge

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1 2

where


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rThe right path length, rpl(X), of any node X is the length of the pathfollowing right links to a leaf.


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rpl(X) = rpl(right(X))) + 1rpl() = 1

Hrpl(H) = 2


Theorem. (Theorem 6.2 in textbook, restated.)If H is a leftist tree, then size(H) 2 – 1, where r = rpl(H) + 1.

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Corollary.The number ofnodes on the rightpath of a leftist tree with N nodesis at most log(N + 1).

N = size(H)



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N = size(H)Proof of the Corollary:By the theorem, N 2 – 1, thus . . .

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N = size(H)Proof of the Corollary:By the theorem, N 2 – 1, thus N + 1 2 and . . .

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N = size(H)Proof of the Corollary:By the theorem, N 2 – 1, thus N + 1 2 and log(N + 1) r,

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N = size(H)Proof of the Corollary:By the theorem, N 2 – 1, thus N + 1 2 and log(N + 1) r,so log(N + 1) r.

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N = size(H)Proof of the Corollary:By the theorem, N 2 – 1, thus N + 1 2 and log(N + 1) r,so log(N + 1) r. QED

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Proof of the theorem: By induction on r.



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Basis case. r = 1. In this case the tree has at least one node; i.e., size(H) 1 = 2 – 1. 1



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Inductive step. Consider a leftist tree H' withr + 1 nodes on its right path.



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Inductive step. Consider a leftist tree H' withr + 1 nodes on its right path. (We will showthat size(H') 2 – 1.)r+1


Inductive step. Consider a leftist tree H' withr + 1 nodes on its right path. Then the roothas a right subtree with r nodes on the rightpath, and a left subtree with at least r nodes on its right path — why?

H1H

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Inductive step. Consider a leftist tree H' withr + 1 nodes on its right path. Then the roothas a right subtree with r nodes on the rightpath, and a left subtree with at least r nodes on its right path (otherwise H' would not be leftist).

H1H

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Applying the inductive hypothesis, and using thenotation in the diagram, . . .

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Applying the inductive hypothesis, and using thenotation in the diagram, size(H ) 2 – 1, size(H) 2 – 1,

H1H

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1 r

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Applying the inductive hypothesis, and using thenotation in the diagram, size(H ) 2 – 1, size(H) 2 – 1,and therefore size(H' ) 2 – 1.

H1H

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r + 1


Applying the inductive hypothesis, and using thenotation in the diagram, size(H ) 2 – 1, size(H) 2 – 1,and therefore size(H' ) 2 – 1. QED

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Merge algorithm, applied to:

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H1

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Merge algorithm: If either of the two heaps is empty, we can return the other.

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Merge algorithm: In this case, neither is empty.

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Merge algorithm: In this case, neither is empty, so we compare their roots. . .

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Merge algorithm: . . . and recursively merge the heap with the largerroot . . .

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Merge algorithm: . . . with the right subheap of the other heap.

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Merge algorithm: . . . producing:

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Merge algorithm: . . . The resulting tree satisfies the heap order property . . .

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Merge algorithm: . . . but is it leftist? . . .

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Merge algorithm: . . . but is it leftist? No, the left subtree has a null path length of 1 while the right subtree has a null path length of 2.

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Merge algorithm: . . . so the algorithm swaps the left subtree with the right . . .

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Merge algorithm: . . . producing:

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Merge algorithm: . . . The result is then a leftist heap, and the algorithm terminates. 3

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T (H , H ) =

O(max(r , r ))


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r = rpl(H ) + 1ii

merge

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where

Again,

This can be proved by induction on r = max(r , r )1 2

T (H , H ) =

O(log (N + 1))


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merge

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where

By the theorem and its corollary,

N = max(N , N )

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H

What about insert and delete_min?

insert( , ) = ?, v


H

insert( , v ) =

H

merge( , ) v

A one element heap


H

insert( , v ) =

H

merge( , ) v

A one element heap

T (H, v) = O(log(N + 1)), where N = size(H).insert


delete_min( ) = ?

H1

H2


delete_min( ) = merge(

H1

H2

H1

H2

, )

T (H) = ? delete_min

6 8 9 11 13 11 Heap: A Special Kind of Tree 15 26 16 25.

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