This document discusses various data structures including heap trees, heap sort, Huffman algorithm, multiway search trees, B trees, and graphs. It provides overviews and examples of heap trees, max heap and min heap insertion and deletion, and the heap sort algorithm. It also gives a brief overview of the Huffman algorithm for data compression and its use of prefix codes. The document is presented by the instructor Mr. S. Christalin Nelson.

1. Instructor
Mr. S.Christalin Nelson
AP(SG)/SoCSE
ADVANCED DATA STRUCTURES
(Vol-2)

2. At a Glance
• Heap Trees
• Heap Sort
• Huffman Algorithm
• Multiway Search Trees
• B Trees
• Graphs
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Data Structures
Instructor: Mr.S.Christalin Nelson
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4. Overview (1/2)
• Heap Tree is a Complete Binary Tree (CBT) whose elements
have keys/values that satisfy the heap property.
• Heap Property
– The keys along any path from root to leaf is ordered either in a
increasing (Max Heap Tree) or decreasing manner (Min Heap
Tree)
• Types
– Min Heap Tree
• Value [root node] ≤ Value [both children]
– Max Heap Tree
• Value [root node] ≥ Value [both children]
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5. Overview (2/2)
• Example
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Instructor: Mr.S.Christalin Nelson
77
55
66
44 50
60 33
22 25
58
18
7
15
10
19 17 16
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6. Max Heap Insertion (1/7)
• Procedure
– (1) Create a new child node and store the new value to it.
– (2) Connect the child to the parent satisfying the CBT property.
– (3) Compare the value of this child with its parent.
• If value[child] > value[parent], then swap the two nodes and go
to step 4.
• If value[child] < value[parent], go to step 5.
– (4) Repeat step 3 until value[child] < value[parent]
– (5) Insertion is complete
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7. Max Heap Insertion (2/7)
• Example: Insert 35, 33, 42, 10, 14, 19, 27, 44, 26, 31, 2 into a
Binary Heap Tree. Delete 2, 44, 27 and show the final Max
Heap Tree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
35
42
33
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8. Max Heap Insertion (3/7)
– Swap 42 & 35.
– Insert 10, 14, 19, 27, 44
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Instructor: Mr.S.Christalin Nelson
42
35
33
10 27
14 19
44
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9. Max Heap Insertion (4/7)
– Swap 44 & 10.
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42
35
33
44 27
14 19
10
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10. Max Heap Insertion (5/7)
– Swap 33 & 44.
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42
35
44
33 27
14 19
10
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11. Max Heap Insertion (6/7)
– Swap 44 & 42.
– Insert 26, 31
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Data Structures
Instructor: Mr.S.Christalin Nelson
44
35
42
33 27
14 19
10 26 31
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12. Max Heap Insertion (7/7)
• Swap 31 & 14. Insert 2
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Instructor: Mr.S.Christalin Nelson
44
35
42
33 27
31 19
10 26 14
• Insertion Complete
2
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13. Max Heap Deletion (1/8)
• Delete 2 (Last leaf node in the CBT)
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Data Structures
Instructor: Mr.S.Christalin Nelson
44
35
42
33 27
31 19
10 26 14 2
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14. Max Heap Deletion (2/8)
• Remove 2
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Data Structures
Instructor: Mr.S.Christalin Nelson
44
35
42
33 27
31 19
10 26 14
• Delete 44 (Root node of CBT)
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15. Max Heap Deletion (4/8)
• Swap 44 (root node) with 14 (last leaf node in the CBT)
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Data Structures
Instructor: Mr.S.Christalin Nelson
14
35
42
33 27
31 19
10 26 44
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16. Max Heap Deletion (4/8)
• Remove 44.
• Compare 14 (present root) with 42 & 35.
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Instructor: Mr.S.Christalin Nelson
14
35
42
33 27
31 19
10 26
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17. Max Heap Deletion (5/8)
• Swap 14 with 42 (biggest value).
• Compare 14 with 33 and 31.
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Data Structures
Instructor: Mr.S.Christalin Nelson
42
35
14
33 27
31 19
10 26
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18. Max Heap Deletion (6/8)
• Swap 14 with 33 (biggest value).
• Compare 14 with 10 and 26.
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Data Structures
Instructor: Mr.S.Christalin Nelson
42
35
33
14 27
31 19
10 26
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19. Max Heap Deletion (7/8)
• Swap 14 with 26 (biggest value).
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Data Structures
Instructor: Mr.S.Christalin Nelson
42
35
33
26 27
31 19
10 14
• Delete 27
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20. Max Heap Deletion (8/8)
• Swap 27 with 14 (last leaf node in CBT). Remove 27
• Compare 14 with 35. No change.
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Instructor: Mr.S.Christalin Nelson
42
35
33
26 14
31 19
10
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21. Exercise
• Q1: Insert 13, 14, 15, 18, 11, 12, 17, 16 into a Binary Max
Heap Tree & Min Heap Tree.
• Q2: The Breadth-first traversal pattern of a Binary tree is 8,
3, 4, 1, 5, 7, 9, 2, 6, 0. Convert the tree into a Binary Max
Heap Tree
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Data Structures
Instructor: Mr.S.Christalin Nelson
• A1:
– Binary Max Heap Tree: 18, 16, 17, 15, 11, 12, 14, 13
– Min Heap Tree: 11, 13, 12, 16, 14, 15, 17, 18
• A2:
– Binary Max Heap Tree: 9, 6, 8, 3, 5, 7, 4, 2, 1, 0
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23. Heap Sort (1/18)
• The root node of a Heap Tree is either the largest or smallest
element.
• Process
– (1) Construct Binary heap tree (max/min).
– (2) Remove the root node and store it in an array and replace
the rightmost leaf node as the new root.
– (3) Repeat steps 1 & 2 until tree is made empty. The final array
consists of sorted elements.
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24. Heap Sort (2/18)
• Example: The Breadth-First Traversal pattern of a CBT is 8, 3,
4, 1, 5, 7, 9, 2, 6, 0. Sort the elements in the CBT using Heap
Sort Algo.
– Given CBT is
– Lets convert given CBT into a Binary Max Heap Tree
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Data Structures
Instructor: Mr.S.Christalin Nelson
8
4
3
1 9
5 7
2 6 0
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25. Heap Sort (3/18)
• Starting from the rightmost subtree in bottom-most level
check if Max Heap Property is satisfied.
• If Max Heap Property is not satisfied then rearrange the
elements by heapification else check next subtree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
8
4
3
1 9
5 7
2 6 0
8
4
3
6 9
5 7
2 1 0
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26. Heap Sort (4/18)
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Instructor: Mr.S.Christalin Nelson
8
4
3
6 9
5 7
2 1 0
8
9
3
6 4
5 7
2 1 0
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27. Heap Sort (5/18)
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Data Structures
Instructor: Mr.S.Christalin Nelson
8
9
3
6 4
5 7
2 1 0
8
9
6
3 4
5 7
2 1 0
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28. Heap Sort (6/18)
• Binary Max Heap Tree is constructed
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Data Structures
Instructor: Mr.S.Christalin Nelson
8
9
6
3 4
5 7
2 1 0
9
8
6
3 4
5 7
2 1 0
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29. Heap Sort (7/18)
• Remove the root node. The last leaf node is shifted/made
the new root node of the CBT.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
9
8
6
3 4
5 7
2 1 0
0
8
6
3 4
5 7
2 1
• Sorted Array {9}
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30. Heap Sort (8/18)
• Convert CBT to binary Max Heap Tree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
0
8
6
3 4
5 7
2 1
8
7
6
3 4
5 0
2 1
• Sorted Array {9}
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31. Heap Sort (9/18)
• Remove the root node. The last leaf node is shifted/made
the new root node of the CBT.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
8
7
6
3 4
5 0
2 1
1
7
6
3 4
5 0
2
• Sorted Array {9, 8}
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32. Heap Sort (10/18)
• Convert CBT to binary Max Heap Tree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
1
7
6
3 4
5 0
2
7
4
6
3 1
5 0
2
• Sorted Array {9, 8}
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33. Heap Sort (11/18)
• Remove the root node. The last leaf node is shifted/made
the new root node of the CBT.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
7
4
6
3 1
5 0
2
2
4
6
3 1
5 0
• Sorted Array {9, 8, 7}
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34. Heap Sort (12/18)
• Convert CBT to binary Max Heap Tree.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
2
4
6
3 1
5 0
6
4
5
3 1
2 0
• Sorted Array {9, 8, 7, 6}
1
4
5
3 2 0
• Remove the root node. The last leaf node is
shifted/made the new root node of the CBT.
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35. Heap Sort (13/18)
• Convert CBT to binary Max Heap Tree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
1
4
5
3 2 0
5
4
3
1 2 0
• Sorted Array {9, 8, 7, 6, 5}
0
4
3
1 2
• Remove the root node. The last leaf node is
shifted/made the new root node of the CBT.
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36. Heap Sort (14/18)
• Convert CBT to binary Max Heap Tree.
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Data Structures
Instructor: Mr.S.Christalin Nelson
0
4
3
1 2
4
0
3
1 2
• Sorted Array {9, 8, 7, 6, 5, 4}
2
0
3
1
• Remove the root node. The last leaf node is
shifted/made the new root node of the CBT.
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37. Heap Sort (15/18)
• Convert CBT to binary Max Heap Tree.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
2
0
3
1
3
0
2
1
• Sorted Array {9, 8, 7, 6, 5, 4, 3}
1
0
2
• Remove the root node. The last leaf node is
shifted/made the new root node of the CBT.
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38. Heap Sort (16/18)
• Convert CBT to binary Max Heap Tree.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
1
0
2
2
0
1
• Sorted Array {9, 8, 7, 6, 5, 4, 3, 2}
0
1
• Remove the root node. The last leaf node is
shifted/made the new root node of the CBT.
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39. Heap Sort (17/18)
• Convert CBT to binary Max Heap Tree. Remove the root
node. The last leaf node is shifted/made the new root node
of the CBT.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
0
1
1
0
• Sorted Array {9, 8, 7, 6, 5, 4, 3, 2, 1}
0
• The CBT is a Singleton Tree. The root node is copied to array
and the Final Sorted Array {9, 8, 7, 6, 5, 4, 3, 2, 1, 0}
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40. Heap Sort (18/18)
• Note:
– Alternatively, instead of removing root node from heap tree, it
can be swapped with the rightmost leaf and retained in the
tree. This node should not be considered for further
operations until the sorting procedure is completed.
– Then, the breadth-first traversal pattern would be 0, 1, 2, 3, 4,
5, 6, 7, 8, 9
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Data Structures
Instructor: Mr.S.Christalin Nelson
0
2
1
3 6
4 5
7 8 9
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42. Overview (1/2)
• Lossless data compression algorithm that assigns a variable-
length unique code (also called bit string or binary string or
code or codeword) to each symbol or character in the input
data based on its frequency of occurrence.
– i.e. The most frequent symbol has the shortest code.
– E.g. Consider a string “ACCEBFFFFAAXXBLKE”
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Character Frequency Code
A 3 111
B 2 100
C 2 101
E 2 001
F 4 01
K 1 0000
L 1 0001
X 2 110
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43. Overview (2/2)
• Huffman code is a prefix code (or) prefix-free code
– i.e. The code representing a particular symbol is never a prefix
of the code representing any other symbol
– Example
• {01, 10, 0010, 1111} is prefix free
• {01, 10, 0010, 1010} is not prefix free because 10 is a prefix of
1010.
– Every message encoded by a prefix free code is uniquely
decipherable
• Variable-Length (prefix) code uses the least space when
compared to the Fixed-length code
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Data Structures
Instructor: Mr.S.Christalin Nelson
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44. Activity (1/2)
• (Q1) Suppose we want to store messages made up of 4
characters a, b, c, d with frequencies 60, 5, 30, 5 (percents)
respectively. What are the fixed-length codes and prefix-free
codes that use the least space?
– To store 100 of these characters,
• Fixed-length code requires 100×2 = 200 bits
• Prefix code requires (60×1) + (5×3) + (30×2) + (5×3) = 150 bits
– Prefix-free codes use least space & offers a 25% saving.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Characters a b c d
Frequency 60 5 30 5
Fixed-Length Code 00 10 10 11
Variable-length code (prefix code) 0 110 10 111
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45. Activity (2/2)
• (Q2) Code is {a=0, b=110, c=10, d=111}. Check if the message
“01101100” is uniquely decipherable.
– 01101100 => 01101100 => 01101100 => 01101100 => “abba”
• The given codes are of variable length & are prefix-free codes. So,
the given message is uniquely decipherable to “abba”. Hence,
{a=0, b=110, c=10, d=111} is a Huffman code.
• (Q3) Code is {a=1, b=110, c=10, d=111}. Check if the message
“1101111” is uniquely decipherable.
– 1101111 => 1101111 => “acad”
– 1101111 => 1101111 => “bad”
• Though the given codes are of variable length, still they are not
prefix-free codes. So, the given message is not uniquely
decipherable as we get two messages “acad” and “bad”. Hence,
{a=1, b=110, c=10, d=111} is not a Huffman code.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
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46. Huffman Algorithm (1/15)
• Huffman Algorithm is used to find the Huffman code
through construction of a Huffman Tree.
• Procedure
– (1) Create a leaf node for each unique character and build a
min heap of all leaf nodes.
– (2) Extract 2 nodes with the minimum frequency from the min
heap
– (3) Create a new internal node with frequency equal to the
sum of the 2 node frequencies. Make the first extracted node
as its left child and the other extracted node as its right child.
Add this node to the min heap.
– (4) Repeat steps 2&3 until the heap contains only one node.
The remaining node is the root node and the tree is complete.
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Data Structures
Instructor: Mr.S.Christalin Nelson
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47. Huffman Algorithm (2/15)
• (Q) Suppose we want to store messages made up of 6
characters a, b, c, d, e, f with frequencies 5, 9, 12, 13, 16, 45
respectively. Find the Huffman code.
– Arrange the characters in ascending order of its frequencies
– Construct a min-heap tree
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Characters a b c d e f
Frequency 5 9 12 13 16 45
a/5
b/9 c/12
d/13 e/16 f/45
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48. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
b/9
d/13 c/12
f/45 e/16
Remove 9
e/16
d/13 c/12
f/45
To min-heap
c/12
d/13 e/16
f/45
a/5
b/9 c/12
d/13 e/16 f/45
f/45
b/9 c/12
d/13 e/16
b/9
d/13 c/12
f/45 e/16
Remove 5 To min-heap
Extract 2 nodes with smallest frequency.
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49. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Create new internal node (N1)
• Frequency (N1) = Sum of freq. of the 2 extracted nodes = 5+9 = 14
• Create a subtree with node (N1) as root, 1st extracted node as left child
& 2nd extracted node as right child.
• Insert N1 into the min heap tree
Characters a b c d e f
Frequency 5 9 12 13 16 45
N1/14
b/9
a/5
c/12
d/13 e/16
f/45 N1/14
c/12
d/13 e/16
f/45
Insert N1
It’s a Min
Heap Tree
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50. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Remove 13 To min-heap
Remove 12 To min-heap
Extract 2 nodes with smallest frequency.
c/12
d/13 e/16
f/45 N1/14
N1/14
d/13 e/16
f/45
d/13
N1/14 e/16
f/45
d/13
N1/14 e/16
f/45
f/45
N1/14 e/16
N1/14
f/45 e/16
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51. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Create new internal node (N2)
• Frequency (N2) = Sum of freq. of the 2 extracted nodes = 12+13 = 25
• Create a subtree with node (N2) as root, 1st extracted node as left child
& 2nd extracted node as right child.
• Insert N2 into the min heap tree
Characters a b c d e f
Frequency 5 9 12 13 16 45
N2/25
d/13
c/12
N1/14
f/45 e/16
N2/25
To min-heap
N1/14
N2/25 e/16
f/45
Insert N2
N1/14
f/45 e/16
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52. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Extract 2 nodes with smallest frequency.
N1/14
N2/25 e/16
f/45
Remove 14 f/45
N2/25 e/16
To min-heap e/16
N2/25 f/45
Remove 16 To min-heap
e/16
N2/25 f/45
f/45
N2/25
N2/25
f/45
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53. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Create new internal node (N3)
• Frequency (N3) = Sum of freq. of the 2 extracted nodes = 14+16 = 30
• Create a subtree with node (N3) as root, 1st extracted node as left child
& 2nd extracted node as right child.
• Insert N3 into the min heap tree
Characters a b c d e f
Frequency 5 9 12 13 16 45
N3/30
e/16
N1/14
Insert N3
N2/25
f/45
N2/25
f/45 N3/30
It’s a Min
Heap Tree
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54. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Extract 2 nodes with smallest frequency.
N2/25
f/45 N3/30
Remove 25 Remove 30
N3/30
f/45
It’s a Min
Heap Tree
f/45
It’s a Min
Heap Tree
Create new internal node (N4)
• Frequency (N4) = Sum of freq. of the 2 extracted nodes = 25+30 = 55
• Create a subtree with node (N4) as root, 1st extracted node as left child
& 2nd extracted node as right child.
• Insert N4 into the min heap tree
N4/55
N3/30
N2/25
f/45
Insert N4
f/45
N4/55
It’s a Min
Heap Tree
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55. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Extract 2 nodes with smallest frequency.
f/45
N4/55
Remove 45 Remove 55
It’s a Min
Heap Tree
N4/55
Empty
Create new internal node (N5)
• Frequency (N5) = Sum of freq. of the 2 extracted nodes = 45+55 = 100
• Create a subtree with node (N5) as root, 1st extracted node as left child
& 2nd extracted node as right child.
• Insert N5 into the Empty tree as root node
N5/100
N4/55
f/45
Empty N5/100
Insert N5
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56. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
The existing internal nodes are:
N1/14
b/9
a/5
N2/25
d/13
c/12
N3/30
e/16
N1/14
N4/55
N3/30
N2/25
N5/100
N4/55
f/45
N5/100
Fit internal nodes into tree:
N5/100
f/45 N4/55
N2/25
c/12 d/13
N3/30
N1/14
a/5 b/9
e/16
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57. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
The generated Huffman Tree is:
N5/100
f/45 N4/55
N2/25
c/12 d/13
N3/30
N1/14
a/5 b/9
e/16
It’s a Full Tree
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58. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
For every parent, consider ‘0’ for Left Child traversal & ‘1’ for Right child traversal
N5/100
f/45 N4/55
N2/25
c/12 d/13
N3/30
N1/14
a/5 b/9
e/16
0
0
0
0
0
1
1
1
1
1
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59. 23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Find code for each character by traversal from root
N5/100
f/45 N4/55
N2/25
c/12 d/13
N3/30
N1/14
a/5 b/9
e/16
0
0
0
0
0
1
1
1
1
1
Characters a b c d e f
Frequency 5 9 12 13 16 45
Variable-length code (prefix code) 1100 1101 100 101 111 0
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60. Huffman Algorithm (15/15)
• (Q) Suppose we want to store messages made up of 6
characters a, b, c, d, e with frequencies 20, 15, 5, 15, 45
respectively. Find Huffman tree and Huffman code.
– Huffman Tree
– Huffman Code
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Characters a b c d e
Variable-length code (prefix code) 000 001 010 011 1
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62. m-way search tree (1/3)
• A tree whose nodes can have certain no. of children (≥2) and
satisfy a search property (i.e. like BST) is called multi-way or
m-way search tree.
• Value of ‘m’ specifies
– Upper limit (max. no.) of subtrees/children per node.
– Upper limit (max. no.) of keys stored per node. i.e. (m-1) keys.
• Application of m-way search tree to speedup processing
– ‘m’ can be a very large number
• Each node corresponds to a physical block on disk.
• Moving one node to another (i.e. moving m-1 data items or a
disk block) involves reading a disk block. This process is very slow
when compared to moving around a data structure stored in
memory.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
62 of 124

63. m-way search tree (2/3)
• Properties
– All keys of a node are stored in ascending order.
• Children between keys k1 & k2 contains all keys in range k1 to k2.
– Each node has 1 to (m-1) keys.
• i.e. 2 to m children/subtrees.
– No. of non-empty subtrees = 0 (for a leaf) to 1+(no. of keys).
– Root has at least two subtrees (or store one key) unless it is a
singleton tree.
• Advantages
– m-way search trees give the same advantages to m-way trees
that BST gave to binary trees - they provide fast information
retrieval and update.
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64. m-way search tree (3/3)
• Disadvantage
– M-way search trees have the same problem that BST had - i.e.
they can become unbalanced.
– E.g.: 4-way Search Tree
• Note: B-Tree of order m (m-way search tree) is used instead.
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Data Structures
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66. B-Trees (1/5)
• B-Tree is a self-balancing search tree.
• Vs. other self-balancing search trees (like AVL)
– B-Trees can hold huge data that cannot fit in main memory.
– B-Trees reduce the number of disk accesses.
• Disk read operation is done when a block (set of keys) is
unavailable in main memory. It is a very slow operation.
• Most of the tree operations require O(h) disk accesses.
– Height (h) of B-Trees is kept low by putting maximum possible keys
in a B-Tree node (approx. equal to the disk block size).
• Height of B-tree = 0.5log2n to log2n
– Where, n - no. of children
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67. B-Trees (2/5)
• Defined by
– No. of keys per node (x) & no. of children (y). Hence also called
as a “x-y B-Tree”.
• E.g. In 2-3 B-Tree, every internal node is either a 2 node (has 1
key and 2 children) or 3 node (has 2 keys and 3 children).
• E.g. In 2-4 B-Tree, every internal node is either a 2/3/4 node. Also
called 2-3-4 tree.
– Order (determine the no. of subtrees/children)
• E.g. 2-3 B-Tree is of Order 3, 2-4 B-Tree is of Order 4
– Minimum degree or Branching Factor (t) or outdegree
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68. B-Trees (3/5)
• Properties
– All leaves are at same level (i.e. Balanced).
– All keys of a node are stored in ascending order
• Children between keys k1 & k2 contains all keys in range k1 to k2.
– No. of keys in a node
• At least: Root has min. 1 key, All node (except root): min.(t-1) keys.
• At most: All nodes (including root) contain max. (2t-1) keys.
– Max. no. of subtrees per node = No. of keys per node + 1.
• Vs. BST
– B-Tree grows and shrinks from root which is unlike BST. BSTs
grow downward and also shrink from downward.
– Like other balanced BSTs, time complexity to search, insert and
delete is O(log n).
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69. B-Trees (4/5)
• Example-1
– B-Tree with minimum degree(t)=3
• All leaves are at same level
• No. of keys
– Root has 3 keys (>1)
– Nodes (except root) have at least 2 keys i.e. (t-1) keys
– All nodes (including root) may contain at most (2t-1) keys. i.e. 5
• No. of children/subtrees per node = No. of keys in node + 1.
– No. of children of the root node = 3 + 1 = 4
• All keys of a node are sorted in ascending order.
– Keys of root node 3, 30, 60 are sorted
• Subtree between keys k1 and k2 contains all keys in range from
k1 and k2.
– Between 3 and 30 of root node, the subtree contains the keys 4, 5, 6
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Data Structures
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70. B-Trees (5/5)
• Example-2
– 2-3 B-Tree (or) B-Tree of Order 3
(or) 3-way Tree
• Example-3
– 2-4 B-Tree (or) 2-3-4 Tree
(or) B-Tree of Order 4
(or) 4-way tree
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Data Structures
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71. Insertion (1/12)
• Consider the following 2-4 B-Tree
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13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20
24
26 28 30
33
37 39 40
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72. Insertion (2/12)
• Insert 21
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20
24
26 28 30
33
37 39 40
21
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73. Insertion (3/12)
• Insert 23
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20 21
24
26 28 30
33
37 39 40
24
Shift 24
Save 23
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74. Insertion (4/12)
• Insert 23
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Data Structures
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13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20 21
23 24
26 28 30
33
37 39 40
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75. Insertion (5/12)
• Insert 29
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Data Structures
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13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20 21
23 24
26 28 30
33
37 39 40
26 28 29 30
Node in Level 2
already has 3 keys!!
Split Node &
Insert 29
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76. Insertion (6/12)
• Insert 29
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 35
1 2
4 5 6
9
11 12
14 15
20 21
23 24 33
37 39 40
26 28 29 30
28
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3 children in Level 2
would promote 28 to
Level 1

77. Insertion (7/12)
• Insert 29
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 28 35
1 2
4 5 6
9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
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78. Insertion (8/12)
• Insert 7
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13 22 32
3 8 10 18 25 28 35
1 2
4 5 6
9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
4 5 6 7 78 of 124
Node in Level 2 already
has 3 keys. Split the
node & Insert 7

79. Insertion (9/12)
• Insert 7 – Node in Level 2 has 5 children. Promote 5 to Level
1. Now node in Level 1 already has 3 keys. Hence Split parent
node.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 28 35
1 2 9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
4 5 6 7
3 8 10
4
3 5
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80. Insertion (10/12)
• Insert 7 – Promote 5 to Level 0. Here, node already has 3
keys - split parent node.
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
3 8 10 18 25 28 35
1 2 9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
6 7
3 5 8 10
4
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81. Insertion (11/12)
• Insert 7 – Node in Level 1 has 2 keys and only 2 children.
Promote 5 to Level 0. At Level 0, the root node already has 3
keys - split parent node.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
18 25 28 35
1 2 9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
6 7
3 5 8 10
4
22 32
5 13
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82. Insertion (12/12)
• Insert 7 – Node has 2 keys and only 2 children. Promote 13
by creating new level.
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Data Structures
Instructor: Mr.S.Christalin Nelson
13 22 32
18 25 28 35
1 2 9
11 12
14 15
20 21
23 24
33
37 39 40
26
29 30
6 7
3 5 8 10
4
22 32
5 13
13
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84. Terminologies (1/7)
• Graph is a non-linear Data Structure
• Definition of Graph, G = (V, E)
– Set of Nodes or Vertices (V), Set of Edges (E)
• Number of Vertices = |V|, Number of Edges |E|
• Edges connect two vertices (source & destination)
• Types of Edges
– Based on Direction
• Directed edges
– Represented as ordered pair (V1, V2). Here (V1, V2) ≠ (V2, V1)
• Undirected (Bidirectional) edges
– Represented as unordered pair {V1, V2}. Here {V1, V2} = {V2, V1}
– Based on Weights
• Weighted & Unweighted edges
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85. Terminologies (2/7)
• Directed Graph (Digraph) & Undirected Graphs
– Given graph is an undirected graph as it has undirected edges.
– Undirected graph is defined as G = (V, E)
• where
– V = {V1,V2,V3,V4,V5,V6,V7,V8}
– E = {{V1,V2},{V1,V3},{V1,V4},{V2,V5},{V2,V6},{V3,V7},{V4,V8}, …. }
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86. Terminologies (3/7)
• Weighted & Unweighted Graphs
– All graphs with weighted (cost) edges are weighted graphs.
• Edges of a weighted graph is labelled with their corresponding
weight (cost incurred for taking the edge).
– Edges of an unweighted graph is not labelled and are
considered to have the same weight/cost (=1).
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87. Terminologies (4/7)
• Null Graph
– A graph which contains only isolated vertices
• i.e. Set of edges in a null graph is empty. E = {}
• Multi-edge
– Many edges can be formed from a single vertex.
• Parallel Edges
– Edges have the same end nodes/vertices. i.e. Two vertices may
be connected by more than one edge.
• Multi Graph
– A graph having parallel edges.
• Trivial Graph
– A graph with only one vertex.
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88. Terminologies (5/7)
• Loop (or) Self-Loop
– An edge in a graph which is drawn from a vertex to itself.
• Simple graph
– A graph with no loops and no parallel edges.
– Max. number of edges |E|
• Simple undirected graph = n(n – 1)/2. 'n' vertices.
• Simple digraph = n(n-1), approx. equal to square of n
– Min. no. of edges = 0
– No. of simple graphs = 2(n(n-1)/2)
• Dense and Sparse Graph
• Cyclic Graph
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89. Terminologies (6/7)
• Complete Graph
– A simple graph (G) is said to be complete if every vertex in G is
connected with every other vertex.
– A graph with ‘n’ vertex has n(n-1)/2 edges.
• Regular Graph
– A graph in which all the vertices are of equal degree.
• A regular graph of degree r is a graph with each vertex having the
degree r.
• Every null graph is a regular graph of degree zero & a complete
graph is a regular graph of degree n-1.
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90. Terminologies (7/7)
• Vs. Tree
– Tree is also an example of non-linear Data Structure.
– Tree is a special kind of graph with special rules.
• Tree with n nodes, would have (n-1) edges,
• All nodes are reachable through the root, etc.
– In Tree all nodes are connected to root in diff. breadth & depth.
• Applications
– Graph Theory
– Weighted Undirected Graph: Inter-city road map
– Unweighted Undirected Graph: Social Network
– Weighted Digraph: Intra-city road map
– Unweighted Digraph: www, web crawlers
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91. Graph Representation
• Vertex List & Edge List
• Vertex List & Adjacency Matrix (for Dense Graphs)
• Vertex List & Adjacency List (for Sparse Graphs)
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92. Example
• (Q1) Find the DFS & BFS traversal of the following digraph
starting at node S.
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Data Structures
Instructor: Mr.S.Christalin Nelson
• (A1)
– DFS traversal: S, A, D, B, C
– BFS traversal: S, A, B, C, D
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93. Example - Depth First Traversal (1/9)
• Additional Data Structure Used: Stack
• Starting at node with S
– DFS Sequence: S
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Data Structures
Instructor: Mr.S.Christalin Nelson
S
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94. Example - Depth First Traversal (2/9)
– DFS Sequence: S A
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Instructor: Mr.S.Christalin Nelson
A
S
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95. Example - Depth First Traversal (3/9)
– DFS Sequence: S A D
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Data Structures
Instructor: Mr.S.Christalin Nelson
D
A
S
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96. Example - Depth First Traversal (4/9)
– DFS Sequence: S A D B
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Data Structures
Instructor: Mr.S.Christalin Nelson
B
D
A
S
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97. Example - Depth First Traversal (5/9)
– DFS Sequence: S A D B C
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Data Structures
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C
D
A
S
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98. Example - Depth First Traversal (6/9)
– DFS Sequence: S A D B C
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Data Structures
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D
A
S
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99. Example - Depth First Traversal (7/9)
– DFS Sequence: S A D B C
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Data Structures
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A
S
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100. Example - Depth First Traversal (8/9)
– DFS Sequence: S A D B C
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Data Structures
Instructor: Mr.S.Christalin Nelson
S
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101. Example - Depth First Traversal (9/9)
– Final DFS Sequence: S A D B C
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Data Structures
Instructor: Mr.S.Christalin Nelson
Stack
Empty
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102. Example - Breadth First Traversal (1/9)
• Additional Data Structure Used: Queue
• Starting at node with S
– BFS Sequence: S
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Instructor: Mr.S.Christalin Nelson
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103. Example - Breadth First Traversal (2/9)
– BFS Sequence: S A
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Instructor: Mr.S.Christalin Nelson
A
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104. Example - Breadth First Traversal (3/9)
– BFS Sequence: S A B
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Data Structures
Instructor: Mr.S.Christalin Nelson
A B
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105. Example - Breadth First Traversal (4/9)
– BFS Sequence: S A B C
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Data Structures
Instructor: Mr.S.Christalin Nelson
A B C
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106. Example - Breadth First Traversal (5/9)
– BFS Sequence: S A B C
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Data Structures
Instructor: Mr.S.Christalin Nelson
B C
Remove
A
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107. Example - Breadth First Traversal (6/9)
– BFS Sequence: S A B C D
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Data Structures
Instructor: Mr.S.Christalin Nelson
B C D
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108. Example - Breadth First Traversal (7/9)
– BFS Sequence: S A B C D
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Data Structures
Instructor: Mr.S.Christalin Nelson
C D
Remove
B
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109. Example - Breadth First Traversal (8/9)
– BFS Sequence: S A B C D
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Data Structures
Instructor: Mr.S.Christalin Nelson
D
Remove
C
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110. Example - Breadth First Traversal (9/9)
– Final BFS Sequence: S A B C D
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Queue
Empty
Remove
D
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111. Activity (1/2)
• Q2. Find the DFS & BFS traversal of the following digraph
starting at node S.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
• A2
– DFS traversal: S, A, D, G, E, B, F, C
– BFS traversal: S, A, B, C, D, E, F, G
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112. Activity (2/2)
• Q3. Find the DFS & BFS traversal of the following digraph
starting at node A.
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
• A3
– DFS traversal: A, B, D, C, F, G, E
– BFS traversal: A, B, C, D, E, F, G
Weights are
ignored !!
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113. Spanning Tree (1/2)
• A subgraph or tree that spans all vertices of a connected &
undirected graph
• Maximum no. of Spanning Trees for a graph is V(V-2)
– Here V = no. of vertices in the graph
• Example:
– Find the maximum no. of spanning trees for the given graph.
Also draw them.
• V = 3. Hence the given graph has 3(3-2) = 3 spanning trees.
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Data Structures
Instructor: Mr.S.Christalin Nelson
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114. Spanning Tree (2/2)
• Properties of Spanning Tree
– Maximum no. of Spanning Trees for a graph = V(V-2)
– Spanning tree for a graph G = (V, E) has of V vertices and (V-1)
edges after removal of only (E-V+1) edges.
• Example: Consider a graph with V=4, E=6.
– After removal of 3 edges (E-V+1 = 6-4+1) from graph, a spanning
tree would have 4 vertices & 3 edges (V-1 = 4-1)
– Should be maximally acyclic. i.e. Adding a single edge to a
spanning tree would make it a Acyclic graph.
– Should be minimally connected. i.e. Removal of a single edge
from a spanning tree would give a disconnected graph.
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115. Minimum Cost Spanning Tree
• A Spanning Tree which is made of minimum cost edges & has
a minimum total cost.
• Algorithms
– Kruskal’s Algorithm
– Prim’s Algorithm
• Basic Initial steps
– Remove Self-loops
– Remove Parallel edges (Retain the edge with least weight)
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
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116. Kruskal’s Spanning Tree Algorithm (1/4)
• (Q1) Find the MST using Kruskal’s Spanning Tree Algorithm
– Draw Edge Table (Edge in ascending order of their weight)
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
10
24
12
22
28
25 16
14
18
Edge AB CD FG CF DG EG DE BE AC
Weight 10 12 14 16 18 22 24 25 28
A
B
E
D
C
F
G
2 Edges have same
weight
Take both edges &
consider any order
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117. Kruskal’s Spanning Tree Algorithm (2/4)
• (Q1) Find the MST using Kruskal’s Spanning Tree Algorithm
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
10
24
12
22
28
25 16
14
18
10
12
22
25 16
14
18
24
Edge AB CD FG CF DG EG DE BE AC
Weight 10 12 14 16 18 22 24 25 28
Start from smallest weight edge & keep selecting edges that
does not form any circuit with previously selected edges
A
B
E
D
C
F
G
A
B
E
D
C
F
G
B
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118. Kruskal’s Spanning Tree Algorithm (3/4)
• (Q1) Find the MST using Kruskal’s Spanning Tree Algorithm
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
MST
Cost = 99
10
24
12
22
28
25 16
14
18
10
12
22
25 16
14
A
B
E
D
C
F
G
A
E
D
C
F
G
B
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119. Kruskal’s Spanning Tree Algorithm (4/4)
• (Q2) Find the MST using Kruskal’s Spanning Tree Algorithm
by starting at S
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Cost of MST = 17
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120. Prim’s Spanning Tree Algorithm (1/4)
• (Q1) Find the MST using Prim’s Spanning Tree Algorithm
– Draw Table (V x V)
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
10
24
12
22
28
25 16
14
18
A
E
D
C
F
G
B
A B C D E F G
A 0 10 28 ∞ ∞ ∞ ∞
B 10 0 ∞ ∞ 25 ∞ ∞
C 28 ∞ 0 12 ∞ 16 ∞
D ∞ ∞ 12 0 24 ∞ 18
E ∞ 25 ∞ 24 0 ∞ 22
F ∞ ∞ 16 ∞ ∞ 0 14
G ∞ ∞ ∞ 18 22 14 0
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121. Prim’s Spanning Tree Algorithm (2/4)
• (Q1) Find the MST using Prim’s Spanning Tree Algorithm
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Data Structures
Instructor: Mr.S.Christalin Nelson
10
12
22
25 16
14
A
B
E
D
C
F
G
A B C D E F G
A 0 10 28 ∞ ∞ ∞ ∞
B 10 0 ∞ ∞ 25 ∞ ∞
C 28 ∞ 0 12 ∞ 16 ∞
D ∞ ∞ 12 0 24 ∞ 18
E ∞ 25 ∞ 24 0 ∞ 22
F ∞ ∞ 16 ∞ ∞ 0 14
G ∞ ∞ ∞ 18 22 14 0
18
24
28
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122. Prim’s Spanning Tree Algorithm (3/4)
• (Q1) Find the MST using Prim’s Spanning Tree Algorithm
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Data Structures
Instructor: Mr.S.Christalin Nelson
10
24
12
22
28
25 16
14
18
MST Cost = 99
A
E
D
C
F
G
B
10
12
22
25 16
14
A
B
E
D
C
F
G
18
24
28
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123. Prim’s Spanning Tree Algorithm (4/4)
• (Q2) Find the MST using Prim’s Spanning Tree Algorithm by
starting at node S
23-Apr-22
Data Structures
Instructor: Mr.S.Christalin Nelson
Cost of MST = 17
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