6.006- Introduction to Algorithms · 6.006- Introduction to Algorithms Lecture 4 Prof. Piotr Indyk....
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6.006- Introduction to Algorithms Lecture 4 Prof. Piotr Indyk
Transcript of 6.006- Introduction to Algorithms · 6.006- Introduction to Algorithms Lecture 4 Prof. Piotr Indyk....
6.006- Introduction to Algorithms
Lecture 4 Prof. Piotr Indyk
Lecture Overview
• Review: Binary Search Trees • Importance of being balanced • Balanced BSTs
– AVL trees • definition • rotations, insert
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5
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10
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7
10 5
1 6 12
Binary Search Trees (BSTs)
• Each node x has: – key[x] – Pointers: left[x], right[x], p[x]
• Property: for any node x: – For all nodes y in the left subtree of x:
key[y] ≤ key[x] – For all nodes y in the right subtree of x:
key[y] ≥ key[x]
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12 5
1 6
7
root
leaf
height = 3
The importance of being balanced
h = Θ(log n) h = Θ(n)
for n nodes:
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10
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10 5
1 6 12
Balanced BST Strategy
• Augment every node with some data
• Define a local invariant on data
• Show (prove) that invariant guarantees Θ(log n) height
• Design algorithms to maintain data and the invariant
AVL Trees: Definition
• Data: for every node, maintain its height (“augmentation”)
– Leaves have height 0 – NIL has “height” -1
• Invariant: for every node x, the heights of its left child and right child differ by at most 1
x
k k-1
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10 5
1 6 12
2
0
1 1
0 0
7
10 5
1 6 12
3
1
1 2
0 0
4 0
[Adelson-Velskii and Landis’62]
AVL trees have height Θ(log n)
• Let nh be the minimum number of nodes of an AVL tree of height h
• We have nh ≥ 1+ nh-1 + nh-2 ⇒ nh > 2nh-2 ⇒ nh > 2h/2
⇒ h < 2 lg nh
• The constant “2” can be improved
h-1 h-2
h
Invariant: for every node x, the heights of its left child and right child differ by at most 1
How can we maintain the invariant ?
Rotations
A
B
α β γ
RIGHT-ROTATE(B)
B
A
γ β α
LEFT-ROTATE(A)
Rotations maintain the inorder ordering of keys: • a ∈ α, b ∈ β, c ∈ γ ⇒ a ≤ A ≤ b ≤ B ≤ c.
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2
3
LEFT-ROTATE(1) 1
2
3
Insertions
• Insert new node u as in the simple BST – Can create imbalance
• Work your way up the tree, restoring the balance
• Similar issue/solution when deleting a node
h-2
h
h-1
u
+1
Balancing
• Let x be the lowest “violating” node – We will fix the subtree of x and
move up • Assume the right child of x is
deeper than the left child of x (x is “right-heavy”)
• Scenarios: – Case 1: Right child y of x is
right-heavy – Case 2: Right child y of x is
balanced – Case 3: Right child y of x is
left-heavy
y
x
C B A
Case 1: y is right-heavy
y
x
C B A
LEFT-ROTATE(x)
k-1 k+1
k k-1
x
y
A B C
k-1 k-1
k
Case 2: y is balanced
Same as Case 1
y
x
C B A
LEFT-ROTATE(x)
k-1 k+1
k k
x
y
A B C
k-1 k
k
Case 3: y is left-heavy
Need to do more …
y
x
C B A
LEFT-ROTATE(x)
k-1 k+1
k-1 k
x
y
A B C
k-1 k
k-1
Rotations
A
B
α β γ
RIGHT-ROTATE(B)
B
A
γ β α
LEFT-ROTATE(A)
Rotations maintain the inorder ordering of keys: • a ∈ α, b ∈ β, c ∈ γ ⇒ a ≤ A ≤ b ≤ B ≤ c.
1
2
3
LEFT-ROTATE(1) 1
2
3
Case 3: y is left-heavy
y
x
C
B
A k-1 k+1
k-1 k
D
z
k-1 or
k-2
LEFT-ROTATE(x) RIGHT-ROTATE (y)
k-1
x
z
A B D
k
k-1
y
k-1 or
k-2 C
And we are done!
Conclusions
• Can maintain balanced BSTs in O(log n) time per insertion
• Search etc take O(log n) time
Examples of insert/balancing
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41
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11 5029
26
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2
1
1
!
!
!
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41
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11 5029
26
2
1
!
!
!
1
23
Insert(23) x = 29: left-left case
65
41
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11 5026
23
3
2
1
1
!
!
!
65
41
20
11 50
1
!!
Done Insert(55)
29 !
3
2
26
23
1
29! !
65
41
20
11 50
2
!!
2
26
23
1
29! !
x=65: left-right case
55
1
55
41
20
11 50
1
!!
2
26
23
1
29! !
65
Done3
!
Balanced Search Trees …
• AVL trees (Adelson-Velsii and Landis 1962) • Red-black trees (see CLRS 13) • Splay trees (Sleator and Tarjan 1985) • Scapegoat trees (Galperin and Rivest 1993) • Treaps (Seidel and Aragon 1996) • ….
BST for runway reservation system • R = (37, 41, 46, 49, 56) current landing times
• remove t from the set when a plane lands R = (41, 46, 49, 56)
• add new t to the set if no other landings are scheduled within < 3 minutes from t
• 44 => reject (46 in R) • 53 => ok
• delete, insert, conflict checking take O(h), where h is the height of the tree
37 41 46 49 56time (mins)
now x x x x
49
56 41
37 46
5-1
1 3-1
1 1
53
49
56 41
46
4+1
1+1 2
1 1
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10 5
1 6 12
3
1
1 2
0 0
4 0
And some people like to do nothing
