data structures algorithmssubodh/courses/col106/lectures/... · 2019-09-20 · basic data...
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Data Structures &
Algorithms
Subodh Kumar(subodh@iitd.ac.in, Bharti 422)
Dept of Computer Sc. & Engg.
Basic Data Structures! Primitive types! List
! Global rank access! Access only at a single rank! Array, Vector
! Local relative access! Local update! Linked lists
!2
bitsmemory: sequence of bytes
base type
object
Linked List
ObjectArray
ReferenceArray
Basic Data Structures! Primitive types! List
! Global rank access! Access only at a single rank! Array, Vector
! Local relative access! Local update! Linked lists
! Both! Sequence
!3Linked List
ReferenceArray
Basic Data Structures! Primitive types! List
! Global rank access! Access only at a single rank! Array, Vector
! Local relative access! Local update! Linked lists
! Both! Sequence
! Queue/Deque! Stack
Restrict usage
True or False?! log (nk) = O(n)! (log n)k = O(n)! Average case complexity = O(worst case
complexity)! Average case complexity = o(worst case
complexity)
!5
col106quiz@cse.iitd.ac.inFormat:t,t,t,f
Collection! Bag of Objects
! Need a way to identify objects ! Add a new object! Delete an identified object! Check if an identified object is present
!6
Dictionary
public interface Dictionary<K, V> { public V get(K key); public void put(K key, V value); public V remove(K key); public iterator<V> allvalues(); public iterator<K> allkeys();}
Array Implementation
!7
class Pair<A,B> { A one; B two;}public class ArrayMap<K,V> implements Dictionary<K, V> {
private Vector<Pair<K,V>> bag;public void put(K key, V value):
// Find an empty space and put Pair<K,V> therepublic V get(K key):
// Iterate of map looking for bag[i].one == key// Return bag[found].two
// Etc.}
! index = int i(key)
Hashing
!8
Array
(key,value)
Every key should have a different index.
Another key
key
Collision
! index = int i(key)! hashcode = int h(key)! index i = hashcode % N
Hashing
!9
Array
Every key should have a different hashcodeand then index also?
(key,value)
Every key should have a different index.
Another key
key
Collision
0
N-1
Collision Resolution! Separate chaining
!10
Array
(,) (,)
No. elementsLoad Factor ! =No. slots
Uniformly random keys ⇒Probability of collision ≤ min(!, 1)
= average chain length
Collision Resolution! Separate chaining
! Open Addressing! Linear probing
! Quadratic probing
! Double Hashing
! (Pseudo) Random probing!11
Array
k1, v1
k2, v2
Rehash
h1
h2
h3
until space locatedhi(k)= h(k) + ki
hi(k)= h(k) + ki2
hi(k)= h(k) + random(h(k), i)
hi(k)= h(k) + h’(k)i Index is “%N”Choose Prime N
size: N
Cuckoo Hashing
!12
Array
k1, v1k2, v2Rehash
h1
h2
h3
True or False?! Worst case query time for hash tables with
separate chaining is O(n2)! Worst case query time for hash tables with
linear probing is "(n)! Worst case insertion time for hash tables
with separate chaining is O(1)! Worst case deletion time for hash tables with
separate chaining is O(1)
!13
col106quiz@cse.iitd.ac.inFormat:t,t,t,f
Hash of Integer! h(Integer i)
! [(a0 i + a1)% P ] %N ! P is a large prime > N ! (a0, a1) ∈ [0 .. P-1]
!14
Hash Functions! Middle r-bits! Add them up! Polynomial function
! Cyclic shift and add
! Universal hashing!15
object
Value Value Value Value Value
ha(k) = ∑ aiki %Ni=0
rWith prime a:
r+1 partsh = 0 h += Cyclic Shift(h) + ki
00001110011011101010
Universal Hash Function! Choose prime N! Divide key k into r+1 parts
k0, k1, .. kr s.t. max(k0) < N! Choose all possible hash functions where a0, a1, .. ar , are each in {0 .. N-1}! There are Nr+1 unique hash functions! At hash creation time, randomly choose a
! and let h = ha
! No need to explicitly enumerate the hash functions
!16
ha(k) = ∑ aiki % Ni=0
r
Probability (h(k1) = h(k2)) = 1/N
True or False?! Expected query time for hash tables with
load factor = 0.5 is O(1) for separate chaining.
! Expected query time for hash tables with load factor = 0.5 is O(1) for open addressing.
! Worst case deletion time for a hash table is O(1) if cuckoo hashing (open addressing) is used.
!17
col106quiz@cse.iitd.ac.inFormat:t,t,f
Ordered Keys! Put it in a sorted list
! Insertion sort
!18
(,) (,) (,) (,)
(1,.) (3,.) (5,.) (9,.)8>key?
node p = sentinel;node n = sentinel.next;while ( p = n; n = n.next;}p.insertAfter(p, Pair(k, v))
< < <
Insert 8❌ ❌ ❌ ✔
(8,.)sentinel
n != null &&n.key < k) {
Skip-list
!19
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
(1,.) (5,.) (9,.) (14,.)
1495
(1,.) (9,.) ∞
∞
∞
Skip-list
!20
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
(1,.) (5,.) (9,.) (14,.)
(1,.) (9,.) ∞
∞
∞
Find 10
p n
-∞ nr
n
Skip-list
!21
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
(1,.) (5,.) (9,.) (14,.)
(1,.) (9,.) ∞
∞
∞
-∞sentinel
-∞
-∞
10?
10?
(10,.)Insert 10
Search in a Skip List
!22
search(n, k): node p = n; n = n.next; while (n != null && n.key < k) { p = n; n = n.next; } if( search(p.below, k);
if(n == null) return “Fail”;
search(sentinel, k);
n.key == k) return “answer”; n != null &&
insert(n, k, topinsert node p = n; n = n.next; while (n != null && n.key < k) { p = n; n = n.next; } // Handle duplicate if necessary if(level < h) insertafternode(p, k); insert(p.below, k, topinsert
Insert in a Skip List
!23
if(n == null) return;
topinsert = toss_coins_until_tails(maxtosses);insert(sentinel, k, topinsert
level
, height
, level):
, level-1);
);What if topinsert > height?
What about the below ref?
add below ref
Skip-list: Increase height
!24
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
(1,.) (5,.) (9,.) (14,.)
(1,.) (9,.) ∞
∞
∞
-∞sentinel
-∞
-∞
10?
Insert 10
10?
To Delete:Delete from all levels
Skip-list: Analysis
!25
Prob(level i exists) <= n/2i
Prob(level log n exists) <= n/2log n = n/n
Prob(level klog n exists) <= n/2klog n = n/nk = 1/n(k-1)
height = klog n
⇒
i successive heads in any of n experiments
Expected no. of nodes visited at any level = 2= Expected number of tosses before heads
Skip-list
!26
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
(1,.) (5,.) (9,.) (14,.)
(1,.) (9,.) ∞
∞
∞
-∞sentinel
-∞
-∞
Next to 10Just before 10?
Skip-list: Save Space
!27
(1,.) (3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)
sentinel
May store only references at the upper levels
Skip-list: Array
!28
sentinel esent
(3,.) (5,.) (8,.) (9,.) (11,.) (14,.) (19,.)(1,.)2
Could even store in arrays
1 2 1 1213 0(-∞)3
(+∞)
“Linked” Data structures
!29
class A { B b; C c; int i;}
2001b c
iA a = new A(..)
dsl;k;lk;lkd;lsk;lsk;lskdl;skl;dksl;dks
dklsjkdlsjkljslkdjs dsdkjslkdjs
dsjdkjs djskdjsljdsldkj
kjkjkjkjdkfjdkfjkdjf
dsl;k;lk;lkd;lskdkjkjkjlsk;lskdl;skl;dksl;dksdlkdjfkldjflkdjflkdjsdkjslkdjs
dsjdkjs fdkfjdkjfkdjkdjfkdjfkdjskdjsljdsldkj
} A a;
a
4400b c
i
a
90210b c
i
a
A a2;
“Linked” Data structures
!30
class A { B b; C c; int i; A a; A a2;}
…….b c
i
a a2A a = new A(..)
…….b c
i
a a2
…….b c
i
a a2…….
b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2…….
b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2 …….b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2
…….b c
i
a a2
Doubly Linked List
!31
…….b c
i
a a2…….
b c
i
a a2
At most two nodes refer to any given node, and are reciprocated.
…….b c
i
a a2
Binary Treeclass BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
!32
class BinaryNode<T> { BinaryNode left;
BinaryNode right; T value; methods etc.
}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
class BinaryNode<T> { BinaryNode left; BinaryNode right; T value; methods etc.}
Exactly one other node contains reference to any given node.One special node has no other node with reference to it.
root
Binary tree has two references per node
edge
pathnodenode
Quiz! Keeping skip-list towers in arrays saves
space. What else does it save? [1 word]! How? [maximum 10 words]! What deficiency does it suffer from?
[Maximum 3 words]
!33
col106quiz@cse.iitd.ac.inFormat: time, No separate fetch of below node, fixed count
Binary Tree
!34
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left;
BinaryNode<T> right; T value; methods etc.
}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right; T value; methods etc.}
8
6 10
7 2092
1 4 15
Root
Ordered Tree
parent
child
leaf
Complete TreeProper Tree
Tree
!35
Not Tree?18
8
6 10
7 2092
1 4 15
Root8
6 10
7 2092
1 4 15
Tree xxRoot = null
Root
Tree
!36
8
6 10
7 2092
1 4 15
Root
Subtree
height
SubtreeLeft
RightLeft
= 3BST:left subtree has lowerright subtree has higher
!37
class BinaryNode<T> { BinaryNode<T> left; BinaryNode<T> right;
T value;}
class BinaryTree<T> { BinaryNode<T> root;
int height() { return height(root); } int height(BinaryNode<T> node) { if(node == null) return -1; return(1 + max( height(node.left), height(node.right)));} int count() { return count(root); } int count(BinaryNode<T> node) { if(node == null) return 0; return(1 + count(node.left) + count(node.right));}
}
Tree Operations
BST
!38
8
6 1
7 292
1 4 1
class BST<T> { BinaryNode<T> root;
BinaryNode<T> find(T v) { return find(root, v); } BinaryNode<T> find(BinaryNode<T> node, T v) { if(node == null) return null; if(node.value == v) return node; BinaryNode n = find(node.left, v); return (n == null)? find(node.right, v) : n;}
class BST<T extends Comparable<T>> { BinaryNode<T> root;
BinaryNode<T> find(T v) { return find(root, v); } BinaryNode<T> find(BinaryNode<T> node, T v) { if(node == null) return null;
int compared = v.compareTo(node.value); if(compared == 0) return node; if(compared < 0) return find(node.left, v); return find(node.right, v);}
BST Operations
!39
8
6 10
7 2092
1 4 15
Root
Insert 5
5
BST Operations
!40
8
6 10
7 2092
1 4 15
Root
Remove 10
5 17
1515
BST Operations
!41
8
6 10
7 92
1 4
Root
Remove 10
5
If Removed node has one child: Promote that child
Otherwise: Transplant up the successor;
Promote successor's child
Successor
!42
8
6 10
7 2092
1 4 15
5
Predecessor? Root
True or False?! The maximum height of a binary tree with n
nodes is n-1! The maximum height of a proper binary tree
with n nodes is 2logn + 1! The minimum height of a complete binary
tree with n nodes, is ceil(log (n+1))! The number of non-leaf nodes in an n-node
tree is always <= n/2
!43
Format: t,f,f,fMail: col106quiz@cse.iitd.ac.in
Traversal
!44
8
6 10
7 2092
1 4 15
Root
class BST<T extends Comparable<T>> { BinaryNode<T> root;
void iterate(Consumer<T> op) { iterate(root, op); } void iterate(BinaryNode<T> node, Consumer<T> op) { if(node == null) return; op.accept(node.value); iterate(node.left, op); iterate(node.right, op);}…
} Pre-order Traversal
Traversal
!45
8
6 10
7 2092
1 4 15
Rootinterface SortablePair<K extends Comparable<K>, V> { K key(); V value();}
interface Position2way<T> { Position2way<T> left(); Position2way<T> right(); T value();}
class BST<PT extends SortablePair> { Position2way<PT> root; int numPositions; void iterate(Consumer<PT> op) { iterate(root, op); } void iterate(Position2way<PT> node, Consumer<PT> op) { if(node == null) return; iterate(node.left(), op); op.accept(node.value()); iterate(node.right(), op); }…} In-order Traversal
tree.iterate(t -> {System.out.println(t.value());});
interface SortablePair<K extends Comparable<K>, V> { K key(); V value();}
Traversal
!46
8
6 10
7 2092
1 4 15
Root
!46
class BST<PT extends SortablePair> { Position2way<PT> root; int numPositions; void iterate(Consumer<PT> op) { iterate(root, op); } void iterate(Position2way<PT> node, Consumer<PT> op) { if(node == null) return; iterate(node.left(), op); op.accept(node.value()); iterate(node.right(), op); }…}
Traversal
!47
8
6 10
7 2092
1 4 15
Root
!47
opclass BST<PT extends SortablePair> { Position2way<PT> root; int numPositions; void iterate(Consumer<PT> op) { iterate(root, op); } void iterate(Position2way<PT> node, Consumer<PT> op) { if(node == null) return; iterate(node.left(), op); op.accept(node.value()); iterate(node.right(), op); }…}
Euler’s Tour
Depth First Traversal
if(node.left() != null) iterate(node.left(), op);op.accept(node.value());if(node.left() != null) iterate(node.right(), op);
if(root != null) iterate(root, op);
Traversal
!48
8
6 10
7 2092
1 4 15
Root
861 2 4 7 109 15 20!48
opclass BST<PT extends SortablePair> { Position2way<PT> root; int numPositions; void iterate(Consumer<PT> op) { iterate(root, op); } void iterate(Position2way<PT> node, Consumer<PT> op) { if(node == null) return; iterate(node.left(), op); op.accept(node.value()); iterate(node.right(), op); }…}
In-order Traversal
Breadth-first Traversal
!49
8
6 10
7 2092
1 4 15
Root
86 10 2 7 9 20 1 4 15
q.insert(root)while(q.hasNext()) { x = q.next(); q.insert(x.left); q.insert(x.right); op.accept(x);}
Queue
Depth-first Traversal
!50
8
6 10
7 2092
1 4 15
Root
8 610 27 14
stack.push(root)while(stack.hasany()) { x = stack.pop(); stack.push(x.right); stack.push(x.left); op.accept(x);}
QueueStack
True or False?! The number of internal nodes in a proper
binary tree is less than n/2.! The Euler’s tour of a binary tree enters each
node 3 times. ! The height of a complete binary tree with n
nodes, is ceil(log (n+1)) - 1.! In-order traversal of a binary search tree
operates on the keys of the tree in an increasing order.
!51
Format: f,f,f,fMail: col106quiz@cse.iitd.ac.in
Tree Balance
!52
159
20
10
5
1 4
2
8 Root3
10
9
20
15
41
5
2 8
15
9
20
10
4
1
5
2
8
Balanced Tree
!53
Height balanced: Height of left subtree is within ±1 of that of right subtree
Count balanced (Weight balanced): Weight of left subtree is at least # times the right subtree’s
# < .29
AVL Tree
Rotation
!54
5
2
10
10t
10
-1
AVL Tree Height
!55
159
20
10
5
1 4
2
8 Root3
2 2
0
00
1 0 1
nmin(h) = nmin(h-1) + nmin(h-2) + 1
nmin(0) = 1 nmin(-1) = 0
nmin(h) > 2 nmin(h-2)
h < ⌊1.44 log(n+2)⌋
n > 2(h+.328)*.694 – 2
Rotation
!56
5
2
10
8 15 20
17h h+2h+1
h/h-1 h/h-1
U
C
G
h+1n0
n1
n2n3
h-1 hh-1/h-2
h+2Imbalance on Insertion
Rotation
!57
5
10
15 20
17h h+2
h
U
C
G
h+1n0
n1
n2n3
h-1 h
h+2
20
17 h+2
10U
C
G
h+1
5h
n0
15h
n1n2
n3
hh-1
h+1
Imbalance on Insertion
Rebalancing AVL Tree
!58
10
20
17
10
17
20
20
10
17
20
17
10
GG
C C
2010
17
U.parent().setchild(m, parent.left==U? LEFT:RIGHT);m.left = lo; m.right = hi;lo.right = n1; hi.left = n2;
m
mm
m
hi
lo
lo
hi
lo
hi
lo
hi
m
lohi
n1 n2
n1
n2
n1 n2n1 n2
n1
n2
U UU
U
h
Rotation
!59
5
2
10
8 15 20
17hh-1 h+1
U
C
G
n0
n1
n2n3
h-1/h-2
h/h-1
h+2Imbalance on Deletion
h-1/h-2
/h-1
Rotation
!60
Imbalance on Deletion
h
5
10
15 20
17
h+2
h-1 h+1U
C
G
n0
n1
n2n3
h-1/h-2
h/h-1
h-1/h-2
20
17 h+2
10U
lo
C
Gh
5h-1
n0
15h/h-1
n1n2
n3
hi
mid
h/h+1
/h+1
h-1
h-1
Rotation
!61
Imbalance on Deletion
h-1
5
10
15 20
17
h+2
h-1 h+1U
C
G
n0
n1
n2
h
h-1/h-2 h-1/h-2
n2n3
h-1/h-2 h-1/h-2
Case 2
Rotation
!62
Imbalance on Deletion
h-1
5
10
15 20n3
17
h+2
h-1 h+1U
Cn0
n2
h
G
n1
h-1/h-2 h-1/h-2
17
15 h+2
10U
G
C
5h-1
n0n2
20n1n3
h-1h-1/h-2 h-1/h-2
h h
h+1
Case 2
True or False?! The height of an AVL tree with n nodes is
O(log n)! The difference in the heights of two children
of an AVL tree node is at most 1! The maximum number of nodes requiring re-
structuring when deleting a node in AVL tree with n nodes is log n
!63
Format: t,t,fMail: col106quiz@cse.iitd.ac.in
AVL Updates! On insertion, re-structure the deepest unbalanced
node! C and G lie on the side of the insertion! Reduces the height of the taller side and fixes the
imbalances of all ancestors! On deletion, re-structure the only unbalanced node
! C is on the other side of the deletion! G is the taller child of C
! If both children of C have the same height, choose the Left-left or Right-right configuration for U C G
! Restructure could reduce height causing an imbalance in the parent
! Traverse higher and repeat, until no imbalance remains!64
Common Leaf Level Tree
!65
15
10
20
11
4
1
5
2
7
155
207 1114
10
2
2-3 Tree
2 or 3 children/node
⌊log3(k)⌋ ≤ height ≤ ⌊log2(k)⌋
1 or 2 keys
⌊log3(2n)⌋ ≤ height ≤ ⌊log2(n)⌋For n nodes:
For k keys:
Insert in 2-3 Tree
!66
155
207 111
10
2,
2-3 Tree
4
13
, 14
Insert 12Insert 14
8
9
10
Insert in 2-3 Tree
!67
135
207 11,121
10
2,
2-3 Tree
4
Insert 25
14
25
21,
15,
• Always insert a child with a key‣ At the leaf, the child may be null
• If the insertion causes overflow⇒3 Keys, 4 references‣ Split into two nodes‣ One key, 2 refs each‣ Promote 1 key + 1 node
13 15
11 12
v
x y
• Always insert a child with a key‣ At the leaf, the child may be null
• If the insertion causes overflow⇒3 Keys, 4 references‣ Split into two nodes‣ 1 key, 2 refs each‣ And promote separator
1/2/3 keys per node
Insert in 2-4 Tree
!68
2-4 Tree
30
50
4 5
22 2623
25Works for a-b tree
a > 1
10 15
21 28 49
1 4 7 11 12 14 20
2 5 13
• Start @leaf‣ after replacing deleted key with successor
• Borrow from sibling if it has >1 key‣ Rotate: promote sibling key, demote parent key‣ Graft child
• Otherwise, merge with sibling‣ For parent:‣ demote separator‣ remove 1 reference‣ Fix any underflow
22 2511
Delete in 2-3 Tree
!69
15
1 4 12 14
2 5 13
Delete 10Delete 7
10
7
Delete 20
20
In a-b tree:Merged node must not exceed b childrena-1 children + a children <= b children
21
❌
❌
True/False for a-b tree?! All leaves are at the same depth! Each node has at least b and at most b
children! 2a ≤ b+1! The root is allowed to have fewer than a
children, but no fewer than 2! A node with x children has x-1 keys
!70
Format: t,f,t,t,t/fMail: col106quiz@cse.iitd.ac.in
Binar-izing a 2-4 Tree
!71
28
4921
10 15
21 28 49
10
15 10
15
10
Red Black Tree
Red Black Tree! No red colored node has a red colored child! The number of black colored node on the
path from the root to each null reference is the same
!72
=> height ≤ 2 log (n)
• Null references are assumed black• Root is always black
2-4 tree height: lg4n = .5 lg2n to lg2n [Up to 2 comparisons /node] 2-3 tre height: lg3n = .63 lg2n to lg2n [Up to 2 comparisons /node] AVL tree height between lg2n to 1.4lg2n [Up to 1 comparison /node]
Binar-ized 2-4 Tree
!73
10
15
2
513 21
28
49
1 4 7
11
1214 20 25 5030
10 15
21 28 49
30 50251 4 7 11 1214 20
2 5 13
49
Red-Black Tree Insert
!74
10
15
2
5
1 4 7
13
12
11
14
21
28
20 25 5030
24
22 23
Insert 24Insert 22Insert 23
Insert @null in node n
Red uncle
Color new node RedBlack parent ⇒ DoneRed parent ⇒ Black Uncle: Restructure
Red Uncle: Recolor + Recur up
R-B Tree Deletion
!75
10
15
5
2
1 4 7
13
12
11
14
21 49
28
20 25 5030
Delete 10
24
Expunge node n with @null in child.
Delete 20
Simple if n is red.Also simple if n ’s child is red.
S
P
d
d
S
n1
SP
n2
S
P
n1
n2
Complex Delete Cases
!76
S
P P
d
d-1
n1 n2d d
d
d
n2n1
d
d-1
d
d
S
P
d
n2n1
d-1
d-1
-1
ns
n1 n2 d d
-1-1
Count +1 if color == Red: "if singlechild().color == Red: singlechild.setcolor(black) "putblack();
putblack(): if root(): " if color == Red: setcolor(Black) " s = sibling(); if s.color == Red: Restructure s.child; s.swapcolor(); # if nephew() == !Red: // i.e., null or Black sibling().setcolor(Red); parent.putblack(); else: Restructure sibling.child() "
49
Red-Black Tree
!77
10
15
2
5
1 4 7
13
12
11
14
28
20 25 5030
Expunge node with @null in node n
24
Delete 20
26
21d = 3
Red Sibling ⇒ Restructure P,S,n
P←Red; S←Black
Black Sibling ⇒ No red nephew: Recolor S,P; Fix P. Red nephew: Restructure P,S,n; Done.
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