Post on 20-Dec-2015
CSE 326: Data StructuresAVL Trees
Ben Lerner
Summer 2007
2
The AVL Balance ConditionLeft and right subtrees of every nodehave equal heights differing by at most 1
Define: balance(x) = height(x.left) – height(x.right)
AVL property: –1 balance(x) 1, for every node x
• Ensures small depth– Will prove this by showing that an AVL tree of height
h must have a lot of (i.e. O(2h)) nodes• Easy to maintain
– Using single and double rotations
3
The AVL Tree Data Structure
4
121062
115
8
14137 9
Structural properties
1. Binary tree property
2. Balance property:balance of every node isbetween -1 and 1
Result:
Worst case depth isO(log n)
Ordering property– Same as for BST
15
4
Is this an AVL tree?
2092
155
10
30177
(NULLs have height -1)
We need to be able to:
1. Track the balance
2. Detect imbalance
3. Restore balance
5
111
84
6
3
1171
84
6
2
5
How about these?
10 12
7
6
Height of an AVL Tree
• M(h) = minimum number of nodes in an AVL tree of height h.
• Basis– M(0) = 1, M(1) = 2
• Induction– M(h) = M(h-1) + M(h-2) + 1
• Solution– M(h) > h - 1 ( = (1+5)/2 1.62)
h-1h-2
h
7
Proof that M(h) > h
• Basis: M(0) = 1 > 0 -1, M(1) = 2 > 1-1
• Induction step.M(h) = M(h-1) + M(h-2) + 1 > (h-1 - 1) + (h-2 - 1) + 1 = h-1 + h-2 - 1 = h-2 ( +1) - 1 = h - 1 (Uses fact that 2 = +1)
8
Height of an AVL Tree
• M(h) > h ( 1.62)
• Suppose we have n nodes in an AVL tree of height h.– N > M(h) – N > h - 1
– log(N+1) > h (relatively well balanced tree!!)
9
Node Heights
1
00
2
0
6
4 9
81 5
1
height of node = hbalance factor = hleft-hright
empty height = -1
0
0
2
0
6
4 9
1 5
1
10
Node Heights after Insert 7
2
10
3
0
6
4 9
81 5
1
height of node = hbalance factor = hleft-hright
empty height = -1
1
0
2
0
6
4 9
1 5
1
0
7
0
7
balance factor 1-(-1) = 2
-1
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Insert and Rotation in AVL Trees
• Insert operation may cause balance factor to become 2 or –2 for some node – only nodes on the path from insertion point to
root node have possibly changed in height– So after the Insert, go back up to the root
node by node, updating heights– If a new balance factor (i.e. the difference
hleft-hright) is 2 or –2, adjust tree by rotation around the node
12
Single Rotation in an AVL Tree
2
10
2
0
6
4 9
81 5
1
0
7
0
1
0
2
0
6
4
9
8
1 5
1
0
7
13
Let the node that needs rebalancing be .
There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of .
The rebalancing is performed through four separate rotation algorithms.
Insertions in AVL Trees
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Bad Case #1
Insert(6)
Insert(3)
Insert(1)
Where is AVL property violated?
15
Fix: Apply Single Rotation
3
1 600
16
3
10
1
2
Single Rotation: 1. Rotate between x and child
AVL Property violated at this node (x)
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j
k
X Y
Z
Consider a validAVL subtree
AVL Insertion: Outside Case
h
hh
17
j
k
XY
Z
Inserting into Xdestroys the AVL property at node j
AVL Insertion: Outside Case
h
h+1 h
18
j
k
XY
Z
Do a “rotation from left”
AVL Insertion: Outside Case
h
h+1 h
19
j
k
XY
Z
Single rotation from left
h
h+1 h
20
j
k
X Y Z
“rotation from left” done!(“rotation from right” is mirror symmetric)
Outside Case Completed
AVL property has been restored!
h
h+1
h
21
Single rotation example
21103
205
15
1
2 4
17
21
10
3 20
5
15
1
2
4
17
22
Bad Case #2
Insert(1)
Insert(6)
Insert(3)
Will single rotation fix? (No)(try bringing up Large; doesn’t work)
23
Fix: Apply Double Rotation
3
1 600
1
3
6
1
0
1
2
6
3
1
0
1
2
Balanced?
Intuition: 3 must become root
AVL Property violated at this node (x)
Double Rotation1. Rotate between x’s child and grandchild2. Rotate between x and x’s new child
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j
k
X Y
Z
AVL Insertion: Inside Case
Consider a validAVL subtree
h
hh
25
Inserting into Y destroys theAVL propertyat node j
j
k
XY
Z
AVL Insertion: Inside Case
Does “rotation from left”restore balance?
h
h+1h
26
jk
X
YZ
“Rotation from left”does not restorebalance… now k isout of balance
AVL Insertion: Inside Case
hh+1
h
27
Consider the structureof subtree Y… j
k
XY
Z
AVL Insertion: Inside Case
h
h+1h
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j
k
X
V
Z
W
i
Y = node i andsubtrees V and W
AVL Insertion: Inside Case
h
h+1h
h or h-1
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j
k
X
V
Z
W
i
AVL Insertion: Inside Case
We will do a“double rotation” . . .
1
2
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j
k
X V
ZW
i
Double rotation : first rotation
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j
k
X V
ZW
i
Double rotation : second rotation
32
jk
X V ZW
i
Double rotation : second rotation
right rotation complete
Balance has been restored to the universe
hh h or h-1
33
Double rotation, step 1
104
178
15
3 6
16
5
106
178
15
4
3
16
5
34
Double rotation, step 2
106
178
15
4
3
16
5
10
6 17
8
15
4
3
16
5
35
Imbalance at node X
Single Rotation
1. Rotate between x and child
Double Rotation
1. Rotate between x’s child and grandchild
2. Rotate between x and x’s new child
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Insert into an AVL tree: a b e c d
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9
5
2
11
7
1. single rotation?
2. double rotation?
3. no rotation?
Inserting what integer values would cause the tree to need a:
Single and Double Rotations:
13
30
Student Activity
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Insertion into AVL tree
1. Find spot for new key
2. Hang new node there with this key
3. Search back up the path for imbalance
4. If there is an imbalance:case #1: Perform single rotation and exit
case #2: Perform double rotation and exit
Both rotations keep the subtree height unchanged.Hence only one rotation is sufficient!
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Easy Insert
2092
155
10
3017
Insert(3)
120
0
100
1 2
3
0
Unbalanced?
40
Hard Insert (Bad Case #1)
2092
155
10
3017
Insert(33)
3
121
0
100
2 2
3
00
How to fix?
Unbalanced?
41
Single Rotation
2092
155
10
30173
12
33
1
0
200
2 3
3
10
0
3092
205
10
333
151
0
110
2 2
3
001712
0
42
Hard Insert (Bad Case #2)
Insert(18)
2092
155
10
30173
121
0
100
2 2
3
00
How to fix?
Unbalanced?
43
Single Rotation (oops!)
2092
155
10
30173
121
1
200
2 3
3
00
3092
205
10
3
151
1
020
2 3
3
01712
0
180
180
44
Double Rotation (Step #1)
2092
155
10
30173
121
1
200
2 3
3
00
180
1792
155
10
203
121 200
2 3
3
10
300
180
45
Double Rotation (Step #2)
1792
155
10
203
121 200
2 3
3
10
300
180
2092
175
10
303
151
0
110
2 2
3
0012
018
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Implementation
balance (1,0,-1)
key
rightleft
47
Single Rotation
RotateFromRight(n : reference node pointer) {p : node pointer;p := n.right;n.right := p.left;p.left := n;n := p
}
X
Y Z
n
RotateFromLeft is symmetric
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Double Rotation
• Class participation• Implement Double Rotation in two lines.
DoubleRotateFromRight(n : reference node pointer) {????}
X
n
V W
Z
49
AVL Tree Deletion
• Similar to insertion– Rotations and double rotations needed to
rebalance– Imbalance may propagate upward so that
many rotations may be needed.
50
Arguments for AVL trees:
1. Search is O(log N) since AVL trees are always well balanced.2. The height balancing adds no more than a constant factor to the
speed of insertion, deletion, and find.
Arguments against using AVL trees:1. Difficult to program & debug; more space for height info.2. Asymptotically faster but rebalancing costs time.3. Most large searches are done in database systems on disk and use
other structures (e.g. B-trees).4. May be OK to have O(N) for a single operation if total run time for
many consecutive operations is fast (e.g. Splay trees).
Pros and Cons of AVL Trees
51
Double Rotation Solution
DoubleRotateFromRight(n : reference node pointer) {RotateFromLeft(n.right);RotateFromRight(n);}
X
n
V W
Z