AVL Trees

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Lecture 12

• AVL Trees

Prof. Sin-Min Lee Department of Computer Science
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Types of Trees
trees
dynamic
static
game trees
search trees
priority queues and heaps
graphs
Huffman coding tree
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Complete Binary Tree

• Is a tree with height h where
• Every node is full upto level h-2,
• Level h-1 is completely filled.
• Level h is filled from left to right.
• Yields to array representation. How to compute
  indices of parent and child?
• How many internal nodes and leafs?

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AVL Trees

• Balanced binary search tree offer a O(log n)
  insert and delete.
• But balancing itself costs O(n) in the average
  case.
• In this case, even though delete will be O(log
  n), insert will be O(n).
• Is there any way to have a O(log n) insert too?
• Yes, by almost but not fully balancing the tree
  AVL (Adelson Velskii and Landis) balancing

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Height of a Tree

• Definition is same as level. Height of a tree is
  the length of the longest path from root to some
  leaf node.
• Height of a empty tree is -1.
• Height of a single node tree is 0.
• Recursive definition
• height(t) 0 if number of nodes 1
• -1 if T is empty
• 1 max(height(LT), height(RT))
  otherwise

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AVL Property

• If N is a node in a binary tree, node N has AVL
  property if the heights of the left sub-tree and
  right sub-tree are equal or if they differ by 1.
• Lets look at some examples.

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AVL Tree Example
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Non-AVL Tree
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Transforming into AVL Tree

• Four different transformations are available
  called rotations
• Rotations single right, single left, double
  right, double left
• There is a close relationship between rotations
  and associative law of algebra.

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Transformations

• Single right
• ((T1 T2) T3) (T1 (T2 T3)
• Single left
• (T1 (T2 T3)) ((T1 T2) T3)
• Double right
• ((T1 (T2 T3)) T4) ((T1T2) (T3T4))
• Double left
• (T1 ((T2T3) T4)) ((T1T2) (T3T4))

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Example AVL Tree for Airports

• Consider inserting sequentially ORY, JFK, BRU,
  DUS, ZRX, MEX, ORD, NRT, ARN, GLA, GCM
• Build a binary-search tree
• Build a AVL tree.

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Binary Search Tree for Airport Names
ORY
ZRH
JFK
MEX
BRU
ORD
DUS
ARN
NRT
GLA
GCM
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AVL Balancing Four Rotations
X1
X3
Double right
X2
X2
X1
Single right
X3
X1
X1
X3
X2
Single left
Double left
X2
X1
X3
X1
X2
X3
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An AVL Tree for Airport Names
After insertion of ORY, JFK and BRU
Single right
Not AVL balanced
AVL Balanced
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An AVL Tree for Airport Names (contd.)
After insertion of DUS, ZRH, MEX and ORD
After insertion of NRT?
DUS
MEX
ZRH
ORD
Still AVL Balanced
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An AVL Tree
Not AVL Balanaced
Now add ARN and GLA no need for rotations Then
add GCM
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An AVL Tree
JFK
GCM
ARN
DUS
GLA
Double left
NOT AVL BALANCED
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Search Operation

• For successful search, average number of
  comparisons
• sum of all (path length1) / number of nodes
• For the binary search tree (of airports) it is
• 39/11 3.55
• For the AVL tree (of airports) it is
• 33/11 3.0

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Known Performance Results of AVL trees

• AVL tree is a sub optimal solution.
• How to evaluate its performance?
• Bounds (upper bound and lower bound) for number
  of comparisons
• C gt log(n1) 1
• C lt 1.44 log(n2) - 0.328
• AVL trees are no more than 44 worse than optimal
  trees.

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Proposed by Adelson-Vels Skii G.M. and Landis
Y.M. (1962)
The definition of an AVL tree indicates that the
minimum number of nodes in a tree is determined
by the recurrence equation AVLh
AVLh-1 AVLh-2 1 where AVL0 0 and AVL1 1.
Example If the height is 1, there must be at
least 2 nodes in the AVL
tree.


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• BALANCE FACTORS
• Indicated by the numbers in the nodes
• Equal the height of right subtree minus the
  height of left subtree
• The height of empty node in AVL is -1 otherwise
  it is 0
• All numbers should be 1, 0, or -1

EXAMPLES OF AVL TREES
1
1
-1
-1
-1
0
-1
0
0
0
1
0
0
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1) FIRST CASE -Resulted from inserting a
node in the right subtree of the right child.
Q
0
P
2
P
1
Q
P

1
1
0
Q
0

1







New node inserted
AVL tree is unbalanced.
The AVL tree is now balanced
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2) SECOND CASE -Resulted from inserting a
node into the left subtree of the right
child. (it is more complex.)
P
P
2
1
1
P
Q

-1

-1
Q
Q

0
A
1

-1

R


A



R
R
R
P
0
2
Q
P
R
1
0

2
Q



0


A
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FACTS ABOUT NODE DELETION AND INSERTION

• Require at most 1.44lg(n2) searches.
• Require 1 single or 1 double rotation, depend
  the balance factors
• Deletion can require 1.44lg(n2) rotations in
  the worst case.
• May or may not need rotation at all

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If append a node to a path of all zero
balance factor in AVL tree , no rotations
required except updating the the balance
factors of nodes.
0
-1
0

1


0

-1




0






1



Before updating the nodes
After updating the nodes
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Example of insertion (with rotation) 20 40 60 .
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40
20
40
40
60
20
60
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Example of insertion (with rotation) 20 40 60 10
15
40
40
20
20
20
60
40
60
40
20
10
60
15
40
15
60
10
20
20
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Example of insertion (with rotation) 20 40 60 10
15 12 .
40
40
20
20
20
60
40
60
40
20
10
60
15
40
40
15
40
15
60
15
60
10
10
20
10
20
12
20
60
12
15
15
10
40
11
40
12
20
60
10
12
20
60
11
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Deletion of a node may or may not requires
rotation .
B
0
2
A
1
C
B
A
1

1
0
0


C
Rotation required.
0
0

1


No rotation required

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FACTS ABOUT NODE DELETION AND INSERTION (cont.)

• 22 cases of deletion require rebalancing
• 47 cases of insertion require rebalancing
• Because the more time-consuming deletion
• occurs less frequently than insertion, it will
• not endangering the efficiency of rebalancing
• AVL trees.

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