Data Structures Module 3 Binary Trees Binary Search Trees Tree Traversals AVL Trees B Trees Hash Tables.pptx

Module III
Introduction to Trees and Hash Tables

Syllabus
• Trees - Binary Trees, Binary Search Trees, AVL
Trees, 2-4 trees, Hash tables, tries.
• Binary Trees and Binary Search Trees -
Operations, Representation, Traversals. Hash
tables - concept of hashing, hash function,
collision and collision resolution.

Linear Data Structures
• Arrays, linked lists, stacks and queues were
examples of linear data structures in which
elements are arranged in a linear fashion (ie,
one dimensional representation).

Non-linear Data Structures
• Tree is another very useful data structure in
which elements are appearing in a non-linear
fashion, which requires a two dimensional
representation.

Tree
 Tree is a nonlinear data structure
 The elements appear in a non linear fashion, which
require two dimensional representations.
 Using tree it is easy to organize hierarchical
representation of objects.
 Tree is efficient for maintaining and manipulating
data, where the hierarchy of relationship among the
data is to be preserved.

Tree
 Definition: A tree is a finite set of one or more nodes
such that there is a specially designated node called
the root. The remaining nodes are partitioned into
n>=0 disjoint sets T1, ..., Tn, where each of these sets is
a tree. We call T1, ..., Tn the subtrees of the root.
A
B C
D E F G
H I J K

Tree - Terminologies
 Here there are 11 nodes and 10 edges
 In any tree with N nodes there will be N-1 edges
 Node: Every individual element in a tree is called a node
 Edge/Links: Nodes are connected by using edges
A
B C
D E F G
H I J K

 Root: It is the origin of the tree data structure. In
any tree, there must be only one root node.
A
B C
D E F G
H I
J
K
 Here A is the root node
 In any tree the first node is called root node

 Leaf Node/External Node/Terminal Node:
 The node which does not have a child is
called a leaf node.
 A node without successor is called a leaf node.
A
B C
D E F G
H I J K

 Internal Node/Non-terminal Node:
 An internal node is a node with atleast one child.
 Every non leaf node is called Internal node
A
B C
D E F G
H I J K

 Parent:
 The node which has child/children.
 A node which is predecessor of any other node.
A
B C
D E F G
H I J K
 Here A is the parent of B & C. B is the parent of D, E & F.

 Child: Descendant of any node is called a child
node.
 Here children of A are B & C. Children of B are D, E & F.
A
B C
D E F G
H I J K

Order of
nodes
⚫Children of a node are usually ordered from left-to-
right.
⚫These are two distinct ordered
trees.
a
b c
a
c b

⚫If a and b are siblings, and a is to the left of b, then all
the descendants of a are to the left of all descendants of
b.
⚫A tree in which the order of nodes is ignored is referred
to as unordered tree.

 Sibling: The nodes with the same parent are
called Sibling nodes.
A
B C
D E F G
H I J K

 Degree of the leaf node is
0.
B C
D G
 Degree of a Node: Total number of the children
of the given node
2
A
3 1
2
J
0
K
0
0
2
E
0
F
H
0
I
0

in a given tree
 Here maximum degree is for node B. Its degree is 3.
 So the degree of the given tree is 3
A
B C
G
 Degree of a Tree: It is the maximum degree of
nodes
2
3 1
2
J
0
K
0
0 D
2
E
0
F
H
0
I
0

 Level: In a tree each step from top to bottom is called as
a Level and the Level count starts with '0' and
incremented by one at each level
A
B C
D E F G
H I J K
Level 0
Level 1
Level 2
Level 3

 Depth of a node: It is the total number of edges
from root to that node.
A
B C
D E F G
H
I
 Depth of node E is
J K

 Depth of the tree: It is the total number of edges
from root to leaf in the longest path
A
B C
D E F G
H I J K
 Depth of the tree = 3

 Height of a node: It is the total number of edges
from longest leaf node in its subtree to that node.
A
B C
D E F G
H I J K
 Height of node B is 2

 Height of the tree: It is the total number of edges
from longest path leaf node to
root
A
B C
D E F G
H I J K
 Height of the tree = 3

 Path: Sequence of nodes and edges between two nodes
 Length of the path: Total number of edges in the path
A
B C
D E F G
H I J K
 Path between B and H is B-E-H. Its length is 2
 There is no path between B and G

 The ancestors of a node are all the
nodes along the path from the root to the node.
 The immediate ancestor of a node is the “parent” node
A
B C
D E F G
H
I

J K

 A descendant node of a node is any node in the
path from that node to the leaf node.
 The immediate descendant of a node is the
“child”
node. A
B C
D E F G
H I
J
K
 Descendant nodes of B are D, E, F, H & I

 Subtrees
A
B C
D E F G
H I J K

Tree Representations
1. List Representation
2. Left Child Right Sibling Representation
3. Representation as a degree 2 tree

List Representations
 Root comes first, followed by list of sub-trees
data link 1 link 2 ... link n

Left child Right sibling Representations
 Fixed sized nodes
 Easier to work
 Two link/pointer fields per node
A
B C
D E F G
H I J K
A
B C
D E F G
H I J K
Left_child data Right_Sibling

Representations as degree 2 tree
 Also known as Left child-Right child Tree/
Degree - Two Tree/ Binary Tree
 Simply rotate the right sibling pointers in the
Left- child Right-sibling tree clockwise by 45 degrees

A
B C
D E F G
A
B C
D E F G
H I J K
Tree
H I J
K
Left Child Right Sibling Representation

Tree - Applications
 Storing naturally hierarchical data: Trees are used to
store the data in the hierarchical structure. Example:
Directory structure of a file system
 Organize data: It is used to organize data for efficient
insertion, deletion and searching.
 Used in compression algorithms
 Routing table: The tree data structure is also used to
store the data in routing tables in the routers.
 To implement Heap data structure: It is used to
implement priority queues

Binary Trees
 A Binary Tree is a finite set of nodes that is either
 Empty or
 Consists of root node and two disjoint
binary trees, called, left subtree and right
subtree
A
B C
D E F
A
B
A

Binary Trees
 Any tree can be transformed into binary tree by left
child-right sibling representation
 A tree can never be empty but binary tree can be
 In binary tree a node can have atmost 2 children,
whereas in case of a tree, a node may have any
number of children

Different Binary Trees
 Skewed Binary Tree
 Left Skewed Binary Tree
 Right Skewed Binary Tree
 Complete Binary Tree
 Full/Strictly Binary Tree

Complete Binary Trees
 A complete binary tree is a binary tree in which
every level, except possibly the last, is completely
filled, and the last level has all its nodes to the left
side
A
B C
D E F
H I
G

Full/Strictly Binary Trees
 Every node other than the leaves has two children
A
B C
D E F
H I
G

Binary Tree Representations
1. Array/Sequential/Linear Representation
2. Linked Representation

Binary Tree -Array Representation
 It is a sequential representation
 Use a 1-D array to store the nodes
 Block of memory for an array is allocated
before storing the actual tree in it.
 Once the memory is allocated, the size of
tree is restricted as permitted by the memory.
 Nodes are stored level by level, starting from zero
level
 The root node stored in first memory location

Binary Tree -Array Representation
 Determine the locations of the parent, left child,
and right child of any node i in the binary tree(1≤i
≤n)
 parent(i)
 If i !=1, then parent(i)= floor(i/2)
 if i =1, i is at root and has no parent
 left_child(i)
 left_child(i) = 2i
 No left child
If 2i ≤ n
Otherwis
e
 right_child(i)
 right_child(i) = 2i+1
 No right child
If 2i+1 ≤ n
Otherwise

Binary Tree -Array Representations
A
B C
D E F
G

B C
D E F
1
A
2 3
4 5 7
10
G
A
B C
D E F
G H

A
B
C
D
E
F
G
B C
D E F
1
2
1
A
2 3
4 5 7
G
10
3
4
5
6
7
8
9
10

A
B C
D E G
F

A
B
C
D
E
F
G
B C
D E F
1
2
3
4
5
6
7
1
A
2 3
4 7
F
5 6

A
B
C
D

A
B
C
D
A
B
C
D
1
2
3
4
5
6
7
8
1
2
4
8

 Advantages
 Any node can be accessed from any other node by
calculating the index.
 Data are stored without any pointer
 Programming languages, where dynamic memory
allocation is not possible(like BASIC, FROTRAN),
array representation is only possible
 Good for Complete Binary tree.
 Searching is fast

 Disadvantages
 Good for Complete Binary tree. If it is not a
complete binary tree then, a lot of memory is
wasted
 Dynamic memory allocation is not possible

Binary Tree -Linked Representations
data
Left_chil
d
Right_child
Left_child data Right_child

A
B C
D E F
G
A
B C
F
D E
G
NULL NULL
NULL NULL
NULL
NULL NULL NULL

Binary Tree –Linked Representations
A
B
C
D
A
B
C
D
NULL
NULL
NULL
NULL
NULL

 Advantages
 Dynamic memory allocation is possible

Binary Tree -Properties
 The maximum number of nodes on level i
of a binary tree is 2i, i>=0
A
B C
D E G
F
H I J K L M N O
Level
0
No. of Nodes
1
1 2
2 4
3 8
i
2i

 The maximum number of nodes in a binary
tree of height h is 2h+1-1, h≥0
A
Height
0
No. of Nodes
1

A
B C
Height
0
No. of Nodes
1
1 3

A
B C
D E G
F
Height
0
No. of Nodes
1
1 3
2 7

A
B C
D E G
F
H I J K L M N O
Height
0
No. of Nodes
1
1 3
2 7
3 15

A
B C
D E G
F
H I J K L M N O
Height
0
No. of Nodes
1
1 3
2 7
3 15
h
2h+1-1

 For any nonempty binary tree, T, if n0 is the
number of leaf nodes and n2 the number of nodes
of degree 2, then n0 =n2 +1
A
B C
D E G
F
J
M
 Number of nodes with degree 2 = n2 = 3

 Minimum number of nodes possible in a
binary tree ofheight h is h+1
 Consider this skewed binary tree
 Here, height of the tree = 3
 Minimum number of nodes possible
A
B
C
D

 For any non empty binary tree, if n is the number
of nodes and e is the number of edges then
n=e+1
 Number of nodes = n = 7
 Number of edges = e = 6
A
B C
D E F
G

Binary Tree Traversals
 Traversal means visit each node in the
binary tree exactly once
 There are three commonly used tree traversals
 Inoder Traversal
 Preorder Traversal
 Postorder Traversal

Inorder Traversal
inorder
A
B C
D E F G
D B E A F C
G

Preorder Traversal
preorder
A
B C
D E F G
A B D E C F
G

Postorder Traversal
postorder
A
B C
D E F G
D E B F G C
A

Inorder Traversal
Left  Root  Right
Algorithm Inorder(tree)
1. If tree!= NULL then
1. call Inorder(treelChild)
2. Print treedata
3. call Inorder(treerChild)

Inorder Traversal
A
B C
D E F
G
Inorder Traversal:

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B G

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B G E

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B G E A

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B G E A C

Inorder Traversal
A
B C
D E F
G
Inorder Traversal: D B G E A C F

Inorder Traversal
struct node
{
int data;
struct node *lchild, *rchild;
}
void Inorder ( struct node * root)
{
if (root != NULL)
{
Inorder(root->lchild);
printf(“%d”,root->data);
Inorder(root->rchild);
}
}

⚫H, D, I, B, E, A, J, F, C, K, G

Preorder Traversal
Root  Left  Right
Algorithm Preorder(tree)
2. call Preorder(treelChild)
3. call Preorder(treerChild)

Preorder Traversal
A
B C
D E F
G
Preorder Traversal:

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B D

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B D E

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B D E G

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B D E G C

Preorder Traversal
A
B C
D E F
G
Preorder Traversal: A B D E G C F

Preorder Traversal
struct node
{
int data;
}
void Preorder ( struct node * root)
{
if (root != NULL)
{
printf(“%d”, root ->data);
Preorder(root ->lchild);
Preorder(root ->rchild);
}
}

⚫A, B, D, H, I, E, C, F, J, G, K

Postorder Traversal
Left  Right  Root
Algorithm Postorder(tree)
1. call Postorder(treelChild)
2. call Postorder(treerChild)

Postorder Traversal
A
B C
D E F
G
Postorder Traversal:

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G E

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G E B

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G E B F

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G E B F C

Postorder Traversal
A
B C
D E F
G
Postorder Traversal: D G E B F C A

Postorder Traversal
struct node
{
int data;
}
void Postorder ( struct node * root)
{
if (root != NULL)
{
Postorder(root->lchild);
Postorder(root->rchild);
printf(“%d”,root->data);
}
}

⚫H, I, D, E, B, J, F, K, G, C, A

Binary Search Trees
 Binary search tree(BST) is a binary tree that is empty
or each node satisfies the following properties:
 Every element has a key, and no two elements have
the same key
 The keys in a nonempty left subtree must be
smaller than the key in the root of the subtree. The
keys in a nonempty right subtree must be larger
than the key in the root of the subtree
 The left and right subtrees are also BST

Binary Search Trees
K
Keys <
K
Keys> K

Binary Search Trees
5
3 6
1 4 8
7 9
5
3 6
1 2 9
7 8
Binary Search Tree Not a Binary Search Tree

Binary Search Trees
5
3 6
1 4 8
7 9
Inorder Traversal: 1 3 4 5 6 7 8 9
Inorder traversal of BST generate a sorted list

Operations on Binary Search Trees
 Four operations
 Searching
 Insertion
 Deletion
 Traversal
 The time complexity of searching, insertion
and deletion = O(h)
where h is the height of the tree.

Binary Search Trees: Searching
 Let item is the value to be searched
 Search begins at root
 If root is NULL search unsuccessful
 Else compare item with root
 If item is less than root then only the left sub tree is to be
searched. The sub tree may be searched recursively
• If item is greater than root then only the right sub tree is
to be searched. The sub tree may be searched recursively

30
20
10
65
80
24 50
45 70 90
68
Search 68
66 69

30
20
10
65
80
24 50
45 70 90
68
Search 68
66 69
Search Data Found

30
20
10
65
80
24 50
45 70 90
68
Search 25
66 69

30
20
10
65
80
24 50
45 70 90
68
Search 25
66 69
Search Data Not Found

Algorithm BST_Search(item)
1. ptr=root
2. flag=0
3. While ptr!=NULL and flag=0 do
1. If ptrdata=item then
1. flag=1
2. Else if ptrdata<item then
1. ptr=ptrrChild
3. Else
1. ptr=ptrlChild
4. If flag=1 then
1. Print “Search data found”
5. Else
1. Print “Search data not found”

Binary Search Trees: Insertion
 Suppose „item‟ is to be inserted in the BST
 Search for item
 If item is found then do nothing
 Else the item is inserted as left or right child where
the search halts

Algorithm BST_Insertion(item)
1. If ROOT=NULL then
1. Create a node new and insert item in to it. Set it as ROOT
node
2. Else
1. Search item in the tree
2. If search data found, insertion is not possible
3. Else
1. Find the parent node in which the item is to be inserted
2. Create a node new and insert item in to it.
3. If item<parentdata then
1. Insert new as the left child of parent
4. Else
1. Insert new as the right child of parent

30
5
2
40
25 80
Insert 20

30
5
2
40
25 80
Insert 20
30
5
2
40
25 80
20

Algorithm BST_Insertion(item)
1. If ROOT=NULL then
1. Create a node new
2. newdata=item
3. newlChild=newrChild=NULL
4. ROOT=new
2. Else
1. ptr=ROOT
2. flag=0
1. If item = ptrdata then
1. flag=1
2. Print “item al ready exist”
3. Exit

2. Else If item < ptrdata then
1. parent=ptr
2. ptr=ptrlChild
3. Else if item > ptrdata
1. parent=ptr
2. ptr=ptrrChild
4. If ptr=NULL then
1. Create a node new
2. newdata=item
3. newlChild=newrChild=NULL
4. If parentdata < item then
1. parentrChild=new
5. Else
1. parentlChild=new

Binary Search Trees: Deletion
Algorithm BST_Deletion()
1. Let x be a node to be deleted
2. If x is a leaf node then
1. Delete x
3. Else if x is a node with only one child then
1. Replace x with the child node of x
2. Delete x
4. Else if x having two children then
1. Find the inorder successor/predecessor of
x, say y
2. Find the right/left child of y, say z
3. Copy the data from y to x

Case 1: x is a leaf node
• Delete x
30
5
2
40
80
30
5
2
40
80
15
x
Delete 15
Case 1

Case 2: x is a node with only one
child
• Replace x with the child node of x
• Delete x
30
5
2
40
80
30
5
2
80
x Delete 40
Case 2
70 90
70 90

Case 3: x having two children
• Find the inorder successor, say y
• Find the right, say z
• Copy the data from y to x
• Replace y by z
30
10 80
20 60
24 50
45 70 90
65
68
Delete 60
66 69

10 80
Delete 60
• Replace y by z
30
x
20 60
24 50
45 70 90
65
68
y
66 69
z

10 80
• Replace y by z
30
x
20 65
24 50
45 70 90
65
68
y
66 69
z
Delete 60

10 80
• Replace y by z
30
x
20 65
24 50
45 70 90
66 69
68 z
Delete 60

Algorithm BST_Deletion(item)
1. ptr=ROOT, flag=0
1. If item < ptrdata then
1. parent=ptr
2. ptr=ptrlChild
2. Else if item > ptrdata then
1. parent=ptr
2. ptr=ptrrChild
3. Else
1. flag=1
3. If (flag==0)
1. Print “ITEM does not exist”
2. Exit

4. Else
1. If ptrrChild=NULL and ptrlChild=NULL then
1. If parentlChild=ptr then
1. parentlChild=NULL
2. Else
1. parentrChild=NULL
2. Else if ptrlChild!=NULL and ptrrChild!=NULL then
1. y is inorder successor of ptr
2. p is the parent of y
3. z is the right child of y
4. Copy the data from y to ptr
5. parentlChild=z
6. Dispose(y)

3. Else
1. If parentlChild=ptr then
1. If ptrlChild!=NULL then
1. parentlChild=ptrlChild
2. Else
1. parentlChild=ptrrChild
2. Else
1. If ptrlChild!=NULL then
1. parentrChild=ptrlChild
2. Else
1. parentrChild=ptrrChild

• A BST is constructed by inserting the following
numbers in the order:
60,25,72,15,30,68,101,13,18,47,70,34. The
number of nodes in the left subtree is
________.

Binary Search Trees: Applications
 Remove duplicate values
 Consider a collection of n data items A1,A2, …, AN.
Suppose we want to find and delete all duplicates
in the collection. Then, we can represent them as
BST and remove the duplicate values, while it
occurs.
 Find the smallest data- Traverse the left subtree of BST
 Find the largest data- Traverse the right subtree of BST
 Find a particular data- Traverse the whole BST

Height-balanced Binary
Search Trees
⚫A self-balancing (or height-balanced) binary search
tree is any binary search tree that automatically keeps
its height (number of levels below the root) small after
arbitrary item insertions and deletions.
⚫A balanced tree is a BST whose every node above the
last level has non-empty left and right subtree.
⚫Number of nodes in a complete binary tree of height h is
2h+1 – 1. Hence, a binary tree of n elements is balanced
if:
2h – 1 < n <= 2h+1 - 1

AVL
Trees
⚫Named after Russian Mathematicians: G.M. Adelson-
Velskii and E.M. Landis who discovered them in 1962.
⚫An AVL tree is a binary search tree which has the
following properties:
⚫The sub-trees of every node differ in height by at most
one.
⚫Every sub-tree is an AVL tree.

⚫Implementations of AVL tree insertion rely on adding
an extra attribute, the balance factor to each node.
⚫ Balance factor (bf) = HL - HR
⚫An empty binary tree is an AVL tree.
⚫A non-empty binary tree T is an AVL tree iff :
⚫ | HL - HR | <= 1
⚫For an AVL tree, the balance factor, HL - HR , of a node can
be either 0, 1 or -1.

⚫The balance factor indicates whether the tree is:
⚫left-heavy (the height of the left sub-tree is 1 greater than
the right sub-tree),
⚫balanced (both sub-trees are of the same height) or
⚫right-heavy (the height of the right sub-tree is 1 greater
than the left sub-tree).

+1
0
0
0
0
0
+1
0
0
-1
2
-1
+1
0
0
AVL Tree
AVL Tree
0 Not an AVL
Tree

⚫Insertion is similar to that of a BST.
⚫After inserting a node, it is necessary to check each of
the node's ancestors for consistency with the rules of
AVL.
⚫If after inserting an element, the balance of any tree is
destroyed then a rotation is performed to restore the
balance.
AVL Tree
Insertion

⚫For each node checked, if the balance factor remains
−1, 0, or +1 then no rotations are necessary.
⚫However, if balance factor becomes less than -1 or
greater than +1, the subtree rooted at this node is
unbalanced.

⚫Theorem: When an AVL tree becomes unbalanced after
an insertion, exactly one single or double rotation is
required to balance the tree.
⚫Let A be the root of the unbalanced subtree.
⚫There are four cases which need to be considered.
⚫Left-Left (LL) rotation
⚫Right-Right (RR) rotation
⚫Left-Right (LR) rotation
⚫Right-Left (RL) rotation

⚫Assume that A is the node which is unbalanced after
inserting a new node.
⚫LL Rotation: Inserted node is in the left subtree of
left subtree of node A.
⚫RR Rotation: Inserted node is in the right subtree of
right subtree of node A.
⚫LR Rotation: Inserted node is in the right subtree of
left subtree of node A.
⚫RL Rotation: Inserted node is in the left subtree of
right subtree of node A.

⚫LL Rotation: New element 2 is inserted in the left subtree
of left subtree of A , whose bf becomes +2 after insertion.
⚫To rebalance the tree, it is rotated so as to allow B to be
the root with BL and A to be its left subtree and right child
respectively, and BR and AR to be the left and right subtrees of
A.
6
4
A
B
0
+1
AR
BL
6
4
B
+2
A
AR
BR 2
+1
BL
BR
0

6
4
B
AR
BR
BL 2
0
0 A
0
A.Leftchild = B.rightchild
B.Rightchild = A

6
8
A
B
0
-1
AL
BR
BL
6
⚫RR Rotation: New element 10 is inserted in the right subtree
of right subtree of A , whose bf becomes -2 after insertion.
8
B
-1
AL
10
BL
⚫To rebalance the tree, it is rotated so as to allow B to be
the root with A as its left child and BR as its right subtree, and
AL and BL as the left and right subtrees of A respectively.
BR
-2 A

10
B
AL
A 6
0
8
0
0
A.Rightchild = B.leftchild
B.leftchild = A
BL
BR

• Create an AVL tree by inserting the elements
in the following order:
• 10,5,15,1,8,20,25,30,40,50

⚫Left-Left case and Left-Right case:
⚫If the balance factor of A is 2, then the left subtree
outweighs the right subtree of the given node, and the
balance factor of the left child L must be checked. The right
rotation with A as the root is necessary.
⚫If the balance factor of L is +1, a single right rotation (with A
as the root) is needed (Left-Left case).
⚫If the balance factor of L is -1, two different rotations are
needed. The first rotation is a left rotation with L as the root.
The second is a right rotation with A as the root (Left-Right
case).

⚫Right-Right case and Right-Left case:
⚫If the balance factor of A is -2 then the right subtree
outweighs the left subtree of the given node, and the balance
factor of the right child (R) must be checked. The left
rotation with A as the root is necessary.
⚫If the balance factor of R is -1, a single left rotation (with A
as the root) is needed (Right-Right case).
⚫If the balance factor of R is +1, two different rotations are
needed. The first rotation is a right rotation with R as the
root. The second is a left rotation with A as the root (Right-
Left case).

B.F
(A)
B.F.
(L)
B.F.
(R)
Type of
Rotation
Number
of
Rotations
Rotation
(Root for
Rotation)
+2 +1 LL Single Right (A)
+2 -1 LR Double Left (L) &
Right (A)
-2 -1 RR Single Left (A)
-2 +1 RL Double Right (R) &
Left (A)

• Construct AVL tree for the following data
21,26,30,9,4,14,28,18,15,10,2,3,7

AVL Tree
Deletion
⚫If the node is a leaf or has only one child, remove it.
⚫Otherwise, replace it with either the largest in its left subtree
(inorder predecessor) or the smallest in its right subtree
(inorder successor), and remove that node.
⚫The node that was found as a replacement has at most one
subtree.
⚫After deletion, retrace the path back up the tree (parent of the
replacement) to the root, adjusting the balance factors as
needed.

⚫The retracing can stop if the balance factor becomes −1
or
+1 indicating that the height of that subtree has
remained unchanged.
⚫If the balance factor becomes 0 then the height of the
subtree has decreased by one and the retracing needs to
continue.
⚫If the balance factor becomes −2 or +2 then the subtree
is unbalanced and needs to be rotated to fix it.

Reference
⚫Sanjay Pahuja, ‘A Practical Approach to Data
Structures and Algorithms’, First Ed. 2007.
⚫For Height Balanced Trees: AVL, B-Trees, refer:
⚫ Pg. 292 – 296, 301 – 315

https://www.cse.iitd.ac.in/~mausam/courses/col106/autumn2017/lectures/10-234trees.pdf

2-3-4 Trees or 2-4 Trees
https://www.cs.ucf.edu/courses/cop3530/sum2005/twofour.pdf

https://www.cs.purdue.edu/homes/ayg/CS251/slides/chap13a.pdf

2-4 Trees
https://www.cs.purdue.edu/homes/ayg/CS251/slides/chap13a.pdf

• Construct a B-Tree (2-4 tree) with the
following keys inserted in order.
• 30,99,70,60,40,66,50,53,45,42,88,75,80,85,90
• Assume that in case of a split 2nd
key is send to
parent, 1st
goes to left child, 3rd
, and 4th
keys go
to right child.
• Delete 42,75,80,45,90.

Hash Tables
• A Hash table is defined as a data structure
used to insert, look up, and remove key-value
pairs quickly.
• It operates on the hashing, where each key is
translated by a hash function into a distinct
index in an array.
• The index functions as a storage location for
the matching value.

• There are majorly three components of hashing:
• Key: input to the hash function to determine an
index or location for storage of an item in a data
structure.
• Hash Function: function that maps the key to a
location in an array called a hash table. The index
is known as the hash index.
Most commonly used function:
hash (key) = (key%size)
• Hash Table: Hash table is a data structure that
maps keys to values using a special function called
a hash function. Hash stores the data in an
associative manner in an array where each data
value has its own unique index.

Collision Resolution
• Collision occurs when two or more keys map to the
same hash value.
• There are mainly two methods to handle collision:
• Separate Chaining - Separate linked list for each key
• Open Addressing - All values are in hash table itself.
– Linear Probing
– Quadratic Probing
– Double Hashing

Data Structures Module 3 Binary Trees Binary Search Trees Tree Traversals AVL Trees B Trees Hash Tables.pptx

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