§4 AVL Trees

Target : Speed up searching (with insertion and deletion)

Tool : Binary search trees

root

small large

Problem : Although Tp = O( height ), but the height can be as bad as O( N ).

CHAPTER 4

Trees

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§4 AVL Trees

〖 Example〗 2 binary search trees obtained for the months of the year

Nov

Oct

Sept

May

Mar

June

July

Dec

Aug

Apr

Feb

Jan

July

June

Mar

May

Oct

SeptNov

Jan

Feb

Aug

Apr Dec

Entered from Jan to Dec

A balanced tree

Discussion 1:Discussion 1: What are the aWhat are the average search tiverage search times of these twmes of these two trees?o trees?What if the moWhat if the months are enterenths are entered in alphabeticd in alphabetical order?al order?

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§4 AVL Trees

Adelson-Velskii-Landis (AVL) Trees (1962)Adelson-Velskii-Landis (AVL) Trees (1962)【 Definition 】 An empty binary tree is height balanced. If T is a

nonempty binary tree with TL and TR as its left and right subtree

s, then T is height balanced iff

(1) TL and TR are height balanced, and

(2) | hL hR | 1 where hL and hR are the heights of TL and TR , respectively.

【 Definition 】 The balance factor BF( node ) = hL hR . In an

AVL tree, BF( node ) = 1, 0, or 1.5

82

1 4

3

7

7

82

1 4

3 5

4

5

6

3

2

1 7

The height of an empty tree is defined to be –1.

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§4 AVL Trees

〖 Example〗 Input the months Mar

Mar01

May

May0

Nov

Nov0

1

2

May01

Nov00

21

Mar00

0

The trouble maker Nov is in the right subtree’s right subtree of the trouble finder Mar. Hence it is called an RR rotation.

In general:

A1

B0

BL BR

AL

RR

InsertionRR

Rotation

A2

B1

BL BR

AL

BL

B0

A

AL

BR

0

A is not necessarily the root of the tree

Single rotation

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§4 AVL Trees

Aug

May01

Nov00

21

Mar01

1

Aug0

Apr

Apr0

1

2

2

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Discussion 2:Discussion 2: What can we do What can we do now?now?

§4 AVL Trees

May01

Nov00

21

Aug01

1

1

Mar0

0

Apr0

Jan

Jan0

1

1

2

LR

Rotation

Mar01

May01

21

Aug01

0

1

Jan0

0

Apr0

Nov0

In general:

A1

B0

BL

ARC0

CRCL

LR

InsertionA2

B1

BL

ARC1

CRCL

OR

LR

Rotation

BL AR

C0

A1 or 0

CR

B0 or 1

CL

OR

Double Rotation

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§4 AVL Trees

Dec July

Mar01

May01

21

Aug01

1

1

Jan0

Apr0

Nov0

July0

Dec0

Feb

Feb0

1

1

2

2

Discussion 3:Discussion 3: What can we do What can we do now?now?

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§4 AVL Trees

July0

Dec01

Aug01

21

Jan01

0

1

Feb0

0

Apr0

Mar01

May01

211

Nov0

June Oct Sept

June0

1

1

1

2

Nov0

Dec01

Aug1

21

Feb0

1

July1

Apr0

Mar0

May1

June0

Jan0

Oct0

1

2

1

1

Oct0

Dec01

Aug1

21

Feb0

1

July1

Apr0

Mar0

Nov0

June0

Jan0

May0

Sept0

1

1

1

1

Note: Several bf’s might be changed even if

we don’t need to reconstructthe tree.

Another option is to keep a height field for each node.

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Read the declaration and functions in Figures 4.42 – 4.48

Home work:

p.136 4.16 Create an AVL Tree

p.136 4.22

Double Rotation

§4 AVL Trees

One last question: Obviously we have Tp = O( h )

where h is the height of the tree.But h = ?

Let nh be the minimum number of nodes in a height balanced tree of height h. Then the tree must look like

A

h2 h1

A

h2h1OR nh = nh1 + nh2 + 1

Fibonacci numbers: F0 = 0, F1 = 1, Fi = Fi1 + Fi2 for i > 1

nh = Fh+2 1, for h 0

Fibonacci number theory gives thati

iF

2

51

5

1

)(ln12

51

5

12

nOhn

h

h

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§5 Splay Trees

Target : Any M consecutive tree operations starting from an empty tree take at most O(M log N) time.

Does it mean that every operation takes O(log N) time?No. It means that the

amortized time is O(log N).

So a single operation might still take O(N) time? Then what’s the point?

The bound is weaker.But the effect is the same:

There are no bad input sequences.

But if one node takes O(N) time to access, we can keep accessing it

for M times, can’t we?

Sure we can – that only means that whenever a node is accessed,

it must be moved.

Discussion 4:Discussion 4: How shall we move the How shall we move the node?node?

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k5

Fk4

Ek3

Dk2

Ak1

CB

§5 Splay Trees

k5

Fk4

Ek3

D

k2

BA

k1

C

k5

Fk4

E

k2

BA

k1

k3

DC

k5

F

k4

E

k2

BA

k1

k3

DC

k4

E

k5

F

k2

BA

k1

k3

DC

Does NOT work!

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§5 Splay Trees

An even worse case:

Discussion 5:Discussion 5: Try to Try to InsertInsert 1, 2, ..., 1, 2, ..., NN in increasing order, and in increasing order, and then then FindFind them in the same order. What will them in the same order. What will happen? What is the total access time happen? What is the total access time TT((NN)?)?

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§5 Splay Trees

Try again -- For any nonroot node X , denote its parent by P and grandparent by G :

Case 1: P is the root Rotate X and P

Case 2: P is not the root

Zig-zag G

DP

AX

B C

X

G

C D

P

A B

Double rotation

Zig-zig G

DP

CX

A B

Single rotation

X

AP

BG

DC

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§5 Splay Trees

k5

Fk4

Ek3

Dk2

Ak1

CB

k5

Fk4

Ek1

k3

DC

k2

BA

k1

k2

BA

k4

k5

FE

k3

DC

Splaying not only moves the accessed node to the root, but also roughly

halves the depth of most nodes on the path.

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§5 Splay Trees

Insert: 1, 2, 3, 4, 5, 6, 7

7

1

6

5

4

3

2

7

6

5

4

1

3

2

Find: 1

7

6

1

4

5

3

2

1

6

74

5

3

2

Read the 32-node example

given in Figures 4.52 – 4.60

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Discussion 6:Discussion 6: Then what must we do? Then what must we do? Please complete the algorithm description for Please complete the algorithm description for DeletionDeletion..

§5 Splay Trees

Deletions:

Step 1: Find X ;

Are splay trees really better than AVL trees?

Home work:

p.136 4.23 Access a splay tree

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Other Operations on Binary Search Trees

Sort: List the elements in increasing order

Solution: inorder traversal.

Get Height: Compute the heights of the nodes

Solution: postorder traversal.

Get Depth: Compute the depths of the nodes

Solution: preorder traversal.

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Research Project 1Binary Search Trees (25)

This project requires you to implement operations on unbalanced binary search trees, AVL trees, and splay trees. You are to analyze and compare the performances of a sequence of insertions and deletions on these search tree structures.

Detailed requirements can be downloaded fromhttp://acm.zju.edu.cn/dsaa/

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Research Project 2Population (25)

It is always exciting to see people settling in a new continent. As the head of the population management office, you are supposed to know, at any time, how people are distributed in this continent. The continent is divided into square regions, each has a center with integer coordinates (x, y). Hence all the people coming into that region are considered to be settled at the center position. Given the positions of the corners of a rectangle region, you are supposed to count the number of people living in that region. Note: there are up to 32768 different regions and possibly even more queries.

Detailed requirements can be downloaded fromhttp://acm.zju.edu.cn/dsaa/

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