Transcript Document

AVL-Trees: Motivation
•
Recall our discussion on BSTs
–
The height of a BST depends on the order of
insertion
•
•
–
•
E.g., Insert keys 1, 2, 3, 4, 5, 6, 7 into an empty BST
Problem: Lack of “balance” – Tree becomes highly
asymmetric and degenerates to a linked-list!!
Since all operations take O(h) time, where log N <= h
<= N-1, worst case running time of BST operations
are O(N)
Question: Can we make sure that regardless of
the order of key insertion, the height of the
BST is log(n)? In other words, can we keep the
BST balanced?
1
Height-Balanced Trees
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Many efficient algorithms exist for balancing
BSTs in order to achieve faster running times
for the BST operations
–
–
–
–
Adelson-Velskii and Landis (AVL) trees (1962)
Splay trees and other self-adjusting trees (1978)
B-trees and other multiway search trees (1972)
Red-Black trees (1972)
•
•
Also known as Symmetric Binary B-Trees
Will not be covered in this course
2
AVL Trees: Formal Definition
1. All empty trees are AVL-trees
2. If T is a non-empty binary search tree with
TL and TR as its left and right sub-trees, then
T is an AVL tree iff
1. TL and TR are AVL trees
2. |hL – hR| <= 1, where hL and hR are the heights of TL
and TR respectively
hL
TL
TR
hL+1 or hL-1
3
AVL Trees
•
•
•
•
AVL trees are height-balanced
binary search trees
An AVL Tree
Balance factor of a node =
height(left subtree) - height(right
subtree)
An AVL tree can only have balance
factors of –1, 0, or 1 at every node
6
1
0
1
4
0
7
-1
0
9
Red numbers
are Balance Factors
For every node, heights of left and
right subtree differ by no more
than 1
4
AVL Trees: Examples and Non-Examples
6 0
1
0
5
7 -1
4
9 0
0
0
An AVL Tree
1
0
1
4
0
9 0
4
0
5 0
9 1
4
1 0 5
6 0
6 -1
1
9 2
7 -1
8
Non-AVL Tree
0
0
4
2
10 2
1
8 0
1
0 7
8 0
An AVL Tree
An AVL Tree
6 -1
1
6 0
6 1
9 0
0
Non-AVL Tree
3
4
7 -2
8 -1
1
9 0
Non-AVL Tree
Red numbers are Balance Factors
5
AVL Trees: Implementation
•
To implement AVL-trees, associate a height
with each node “x”
C++ Declaration
x
left
key
height
•
•
right
struct AVLTreeNode{
int key;
int height;
AVLTreeNode left;
AVLTreeNode right;
};
Balance factor (bf) of x = height of left
subtree of x – height of right subtree of x
In an AVL-tree, “bf” can be one of {-1, 0, 1}
6
Good News about AVL Trees
•
•
Can prove: Height of an AVL
An AVL Tree
tree of N nodes is always
Height = O(logN)
0
O(log N)
6
How? Can show:
-1
1
–
–
Height h = 1.44 log(N)
Prove using recurrence relation
for minimum number of nodes
S(h) in an AVL tree of height h:
•
–
S(h) = S(h-1) + S(h-2) + 1
5
0
4
7
9
0
Red numbers
are Balance Factors
Use Fibonacci numbers to get
bound on S(h) bound on height h
7
•
Good and Bad News about AVL Trees
Good News:
–
•
Search takes O(h) = O(logN)
Bad News
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Insert and Delete may cause the tree to be unbalanced!
6
1
0
5
Insert 3
0
7
4
An AVL Tree
6 1
-1
9
2
1
0
0
4
5
7
-1
9
0
3
No longer an AVL Tree
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Restoring Balance in an AVL Tree
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Problem: Insert may cause balance factor to become
2 or –2 for some node on the path from root to
insertion point
•
Idea: After Inserting the new node
1.
Back up towards root updating balance factors along the
access path
2. If Balance Factor of a node = 2 or –2, adjust the tree by
rotation around deepest such node.
6 1
2
1
0
4
5
7
-1
9
0
3
Non- AVL Tree
9
Restoring Balance: Example
6
1
0
5
4
Insert 3
0
7
-1
9
AVL
2
1
0
0
Rotate
6 1
5
7
-1
4
9
3
Not AVL
•
6 0
0
0
0
3
7
4
5 0
-1
9
AVL
After Inserting the new node
1. Back up towards root updating heights along the access path
2. If Balance Factor of a node = 2 or –2, adjust the tree by
rotation around deepest such node.
10
0
AVL Tree Insertion (1)
P
P is the Pivot: bf is 2 or -2
L
A
•
R
B
C
D
Let the node that needs rebalancing be P.
–
–
P is called the pivot node
P is the first node that has a bf of 2 or -2 as we
backup towards the root after an insertion
11
AVL Tree Insertion (2)
P
P is the Pivot: bf is 2 or -2
L
A
•
R
B
C
D
There are 4 cases:
–
Outside Cases (require single rotation) :
−
Inside Cases (require double rotation) :
1. Insertion into left subtree of left child of P (LL Imbalance).
2. Insertion into right subtree of right child of P (RR Imbalan.)
3. Insertion into left subtree of right child of P (RL Imbalance)
4. Insertion into right subtree of left child of P (LR Imbalance)
12
LL Imbalance & Correction
Rotate
2
L
P
0 or 1 L
P
R
A
R
A
B
C
Tree after insertion
D
B
C
D
Tree after LL correction
•
LL Imbalance: We have inserted into the left
subtree of the left child of P (into subtree A)
–
–
bf of P is 2
bf of L is 0 or 1
– Correction: Single rotation to the right around13 P
LL Imbalance Correction Example (1)
1
1
4
Insert(2)
10
1
1 4
20 0
0 3
Backup
10
20 0
0 3
Initial AVLTree
0 2
2
4
Rotate
10
0
20 0
1 3
0 2
1
10
20 0
0 3
0 2
Identified 4
as the pivot
Tree right after
insertion of 2
Now, backup the
tree updating
balance factors
1
4 0
AVL Tree after
LL correction
Classify the type
of imbalance
•
LL Imbalance:
–
–
bf of P(4) is 2
bf of L(3) is 0 or 1
14
LL Imbalance Correction Example (2)
1
10
0 4
0 3
Insert(2)
0 4
20 0
5 0
Initial
AVLTree
0 3
Backup
1 10
2 10
1 4
20 0
1 3
5 0
0 2
0 2
Tree right after
insertion of 2
Now, backup the
tree updating
balance factors
Rotate
20 0
5 0
0 4
1 3
5 0
0 2
Identified 10 as
the pivot
10 0
20 0
AVL Tree after
LL Correction
Classify the type
of imbalance
•
LL Imbalance:
–
–
bf of P(10) is 2
bf of L(4) is 0 or 1
15
RR Imbalance & Correction
Rotate
P
R
-2
P
L
R 0 or -1
D
L
A
B
C
Tree after insertion
C
D
A
B
Tree after RR correction
•
RR Imbalance: We have inserted into the right
subtree of the right child of P (into subtree D)
–
–
bf of P is -2
bf of R is 0 or -1
– Correction: Single rotation to the left around16P
RR Imbalance Correction Example (1)
10 -1
0 4
Insert(40)
20 -1
30 0
Initial
AVLTree
0 4
10 -1
20
Backup
0 4
-1
20
-2
40 0
40 0
Identified 20
as the pivot
Tree right after
insertion of 40
10 -1
30 0
0 4
30 -1
30 0
Now, backup the
tree updating
balance factors
Rotate
10 -1
0 20
40 0
AVL Tree after
RR Correction
Classify the type
of imbalance
•
RR Imbalance:
–
–
bf of P(20) is -2
bf of R(30) is 0 or -1
RR Imbalance Correction Example (2)
10 -1
0 4
0 15
10
Insert(40) &
Backup
20 0
0 4
30 0
Rotate
-2
20 -1
0 15
Initial AVLTree
0 10
30 -1
0 4
30 -1
15 0
40 0
40 0
Tree after
insertion of 40
As we backed up the
tree, we updated
balance factors and
identified 10 as the
pivot
20 0
AVL Tree after
RR Correction
Classify the type
of imbalance
•
RR Imbalance:
–
–
bf of P(10) is -2
bf of R(20) is 0 or -1
LR Imbalance & Correction
2
-1 L
1
Rotate(1)
P
R
2
LR
C
B1
D
B2
Tree after insertion
•
R
2
LR
P
L
L
LR
A
Rotate(2)
P
B2
A
C
R
D
A
B1 B2
B1
Tree after rotate(1)
C
Tree after LR correction
LR Imbalance: We have inserted into the right
subtree of the left child of P (into subtree LR)
–
–
bf of P is 2
bf of L is -1
D
– Correction: Double rotation around L & P
LR Imbalance Correction Example
1
0
0 3
Insert(5) &
Backup
10
4
2
20 0
7 0
Initial AVLTree
As we backed up the
tree, we updated
balance factors and
identified 10 as the
pivot
10
2
-1 4
0 3
Rotate
0
0 4
20 0
1
7 1
7
0 3
5 0
Tree after
insertion of 5
10 -1
5 0
20 0
AVL Tree after
LR Correction
Classify the type
of imbalance
•
LR Imbalance:
–
–
bf of P(10) is 2
bf of L(4) is -1
20
RL Imbalance & Correction
P
-2
2
L
R
RL
A
Rotate(1)
D
C2
A
B
RL
P
RL
C1
C1
C2
R
L
R
Tree after insertion
•
2
L
1
-2
1
B
C1
P
Rotate(2)
D
Tree after Rotate(1)
A
C2
D
B
Tree after RL correction
RL Imbalance: We have inserted into the left
subtree of the right child of P (into subtree RL)
–
–
bf of P is -2
bf of R is 1
– Correction: Double rotation around L & P
RL Imbalance Correction Example
Insert(17)
-1 10
0
4
20 0
0 15
0 4
2
-1 15
30 0
Initial AVLTree
As we backed up the
tree, we updated
balance factors and
identified 10 as the
pivot
Rotate
-2 10
20 1
1
30 0
15 0
1 10
0 4
17 0
Tree after
insertion of 17
20 1
0 17
30 0
AVL Tree after
RL Correction
Classify the type
of imbalance
•
RL Imbalance:
–
–
bf of P(10) is -2
bf of R(20) is 1
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Deletion
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Deletion is similar to insertion
•
First do regular BST deletion keeping track of
the nodes on the path to the deleted node
•
After the node is deleted, simply backup the
tree and update balance factors
−
−
If an imbalance is detected, do the appropriate
rotation to restore the AVL tree property
You may have to do more than one rotation as you
backup the tree
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Deletion Example (1)
1
10
0 4
0 3
Delete(20)
20 0
5 0
Initial AVLTree
1
10
Backup
0 4
0 3
2 10
Rotate
0 4
5 0
Tree after
deletion of 20
Now, backup the
tree updating
balance factors
-1
10 1
0 3
5 0
0 3
4
Idenfitied 10 as
the pivot
0
5
AVL Tree after
LL Correction
Classify the type
of imbalance
•
LL Imbalance:
–
–
bf of P(10) is 2
bf of L(4) is 0 or 1
24
Deletion Example (2)
1
10
-1 4
Delete(20)
20 0
5 0
Initial AVLTree
1
10
Backup
-1 4
2 10
-1 4
1
5 0
Tree after
deletion of 20
Now, backup the
tree updating
balance factors
Rotate
2
5 0
0 4
5
0
10 0
AVL Tree after
LL Correction
Idenfitied 10 as
the pivot
Classify the type
of imbalance
•
LR Imbalance:
–
–
bf of P(10) is 2
bf of L(4) is -1
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Deletion Example (3)
1
1 3
1 2
20
0 1
-1 7
0 4
Delete(10)
10
1 5
1
0 15
0 8
1 5
-1
1 3
30 -1
0 40
1 2
0 1
Initial AVLTree
15
20
-1 7
0 4
0 8
-1
30 -1
0 40
Tree after deletion of 10
We have copied the
successor of 10 to root
and deleted 15
Now, backup the
tree updating
balance factors
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Deletion Example (3) - continued
Backup
1
1 5
1 3
1 2
0 1
20
-1 7
0 4
Rotate
15
-2
0 8
0 40
Identified 20 as the pivot
RR Imbalance:
–
–
bf of P(20) is -2
bf of R(30) is 0 or -1
15
1 5
1 3
30 -1
Classify the type
of imbalance
•
1
1 2
0 1
30 0
-1 7
0 4
20 0 40 0
0 8
Tree after RR imbalance
is corrected
Is this an AVL
tree?
Continue backing
up the tree
updating balance
factors
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Deletion Example (3) - continued
Backup
1 5
1 3
1 2
0 1
Rotate
2 15
30 0
-1 7
0 4
0 5
20 0 40 0
0 8
Identified 15 as the pivot
1 3
1 2
0 1
15 0
0 4
7 -1
30 0
0 8
20 0 40 0
Final Tree
Classify the type
of imbalance
•
LL Imbalance:
–
–
bf of P(15) is 2
bf of L(5) is 0 or 1
28
Search (Find)
•
Since AVL Tree is a BST, search algorithm is
the same as BST search and runs in
guaranteed O(logn) time
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Summary of AVL Trees
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Arguments for using AVL trees:
1. Search/insertion/deletion is O(log N) since AVL
trees are always balanced.
2. The height balancing adds no more than a constant
factor to the speed of insertion/deletion.
•
Arguments against using AVL trees:
1. Requires extra space for balancing factor (height)
2. It may be OK to have a partially balanced tree that
would give performance similar to AVL trees
without requiring the balancing factor
• Splay trees
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