Post on 19-Jan-2016
Data Structures
AVL Trees
Balanced BSTObservation• BST: the shallower the better!• For a BST with n nodes– Average height is O(log n)– Worst case height is O(n)
• Simple cases such as insert(1, 2, 3, ..., n) lead to the worst case scenario: height O(n)
Solution: Require a Balance Condition that1. ensures depth is O(log n)
2. is easy to maintain
– strong enough!
The AVL Tree Data Structure
1. A binary search tree2. Heights of left and
right subtrees of every node differ by at most 1
115
8
Adelson-Velskii and Landis
An AVL tree is:
4
121062
13 147 9
15
Recursive Height CalculationARecall: height is max number of
edges from root to a leaf
hright
How do we compute height at A?
Note: height(null) = -1
hleftt
84
6
AVL Tree?
1 2
3
111
10 12
70
0 0
0 1
84
6
13
4
AVL Tree?
3
111
2
5
0
0 0 7 0
1
2
An AVL Tree has Height O(log n)
8
Proof Let S(h) be the min # of nodes in an AVL tree of height hClaim: S(h) = S(h-1) + S(h-2) + 1
AVL tree of height h=4with the min # of nodes (12)
4
12106
115
15
Solution of recurrence:S(h) = O(2h)(like Fibonacci numbers)
0
11 2
70
20
9 13 14 1
0 0
32
Testing the Balance Property
155
10 We need to be able to:
1. Track Balance
2092
17 307
NULL has a height of -1
2. Detect Imbalance3. Restore Balance
An AVL Tree
2052 2
310
10
3
data
height
children
92
7
0
0 017
030
115
1
Track height at all times.
AVL trees: find, insert, delete• AVL find:– Same as BST find.
• AVL insert:– Same as BST insert, except you need to
check your balance and may need to “fix” the AVL tree after the insert.
• AVL delete:– We’re not going to talk about it, but same
basic idea. Delete it, check your balance, and fix it.
Bad Case #1
Insert(6) Insert(3) Insert(1)
6
3
1
Simple illustration
Fix: Apply Single Rotation
36
1
2 1
AVL Property violated at this node (x)
1 6003
10
Intuition: 3 must become root
Single Rotation:1. Rotate between x and child Simple illustration
Bad Case #2
Insert(1) Insert(6) Insert(3) 6
1
3
Simple illustration
Single Rotation Does Not Work
12
12
AVL Property violated at this node (x)
It’s a BST, but it’s unbalanced
30
1
6
Simple illustration
1
3
0
6
Fix: Apply Double Rotation
311
6
2
3
1
1
2
AVL Property violated at this node (x)
1 600
30
60
Double Rotation1. Rotate between x’s child and grandchild2. Rotate between x and x’s new child Simple
illustration
1
AVL tree insert
1. Find spot for new key2. Hang new node there with this key3. Search back up the path for imbalance4. If there is an imbalance:
Case #1: Perform single rotation and exitCase #2: Perform double rotation and exit
Both rotations keep the subtree height unchanged.Hence only one rotation is sufficient!
Case #1: Single Rotation
h+1
a
ZY
bh
h
X
h+2
h+3
a
b
X < b < Y < a < Z
single rotation
XZY
Single rotation example
4
15
822
3 6
10 19
17 20
24
16
Single rotation example
4
15
822
3 6
10 19
17 20
24
16
Single rotation example
4
15
822
3 6
10 19
17 20
24
16
104
8
15
3 6
19
17
16
22
2420
Case #2: Double Rotation
a
ZX
c
V
h-1
h
h
h
U
h+1
h+2
b
h+3
Let’s break subtree into pieces:
Insert on left child’s right (at U or
V) c
Case #2: Double Rotation
a
ZX
c
V
h-1
h
h
h
U
h+1
h+2
b
h+3
b
Insert on left child’s right (at U or
V)
Let’s break subtree into pieces: c
Case #2: Double Rotation
a
ZX
c
V
h-1
h
h
h
U
h+1
h+3
b
Insert on left child’s right (at U or
V) c
a
XU V
Z
Let’s break subtree into pieces:
h+2
b
Can also do this in two rotations a
Z
b
X
c
V
h-1
h
h
h
U
h+1
h+2h+3
a
Zb
V
h-1h
h h
First rotation
X < b < U < c < V < a < Z
c
Second rotation
a
Zb
c
V
h-1h
h h
U
h+1
h+2
h+3
a
Z
b
X
c
V
h-1 hh h
X
U
25
Second rotation
h+1
Double rotation example
104
8
15
3 6
19
17
2016
22
24
5
Double rotation example
104
8
15
3 6
19
17
2016
22
24
5
Double rotation example
104
8
15
3 6
19
17
2016
22
24
5
Double rotation example
104
8
15
3 6
19
17
2016
22
24
5
15
19
17
2016
22
24
10
8
Double rotation example
104
8
15
3 6
19
17
2016
22
24
5
15
19
17
2016
22
24
10
8
6
4
3 5
Double rotation example 15
19
17
2016
22
24
10
8
6
4
3 5
15
19
17
2016
22
2410
8
6
4
3 5
Case #3:a
X
b
Z
ch-1
hh
h
VU
h+1
h+2
h+3
c
Z
a
X
bh-1 hh h
VU
h+1 h+1
h+2
Double rotation
Case #4:a
X
b
Y
h+1h
h
Z
h+2
h+3
a
X
b
Y
h
h+1
hZ
h+1
h+2
Single rotation
Recap of AVL tree insertLet x be the node where an imbalance occurs.
Four cases to consider. The insertion is in the1. left subtree of the left child of x.2. right subtree of the left child of x.3. left subtree of the right child of x.4. right subtree of the right child of x.
Cases 1 & 4 are solved by a single rotationCases 2 & 3 are solved by a double rotation
AVL complexity
What is the worst case complexity of a find?
O(log n)
What is the worst case complexity of an insert?
O(log n)
What is the worst case complexity of build a AVL Tree?
O(n log n)