two methods on how to construct a min-heap using a sequence od data.

1. DATA STRUCTURES
AND ALGORITHMS
MIN-HEAP
CONSTRUCTION
SADIA ZAR
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W), RAWALPINDI. 1

2. MIN-HEAP – DATA STRUCTURE
• Construct a min-heap from the following sequence:
15, 3, 6, 18 ,5 ,9 ,11
• In this lecture I’ve discussed two methods to construct a min-heap.
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
2

3. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
In the first method, at first we will construct a complete
binary tree first and then we will check the condition i.e
A[Parent[i]] < = A[i] for every node in the tree.
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
3

4. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
4

5. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
5

6. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
6

7. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
7

8. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
8

9. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
9

10. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
10

11. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
11

12. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
i = 3
Parent[i] = Parent[3] = 1
So the condition is A[1] <= A[3]
i.e, 3 <= 18 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
12

13. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
i = 4
Parent[i] = Parent[4] = 1
So the condition is A[1] <= A[4]
i.e, 3 <= 5 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
13

14. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
i = 5
Parent[i] = Parent[5] = 2
So the condition is A[2] <= A[5]
i.e, 6 <= 9 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
14

15. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
i = 6
Parent[i] = Parent[6] = 2
So the condition is A[2] <= A[6]
i.e, 6 <= 11 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
15

16. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
i = 1
Parent[i] = Parent[1] = 0
So the condition is A[0] <= A[1]
i.e, 15 <= 3 (false) so we have to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
16

17. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
After swapping check other nodes as well
i = 2
Parent[i] = Parent[2] = 0
So the condition is A[0] <= A[2]
i.e, 3 <= 6 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
17

18. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
Again check all the nodes now the problem
Lies in node 1 and node 4
i = 4
Parent[i] = Parent[4] = 1
So the condition is A[1] <= A[4]
i.e, 15 <= 5 (false) so we have to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
18

19. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
Now again check
i = 4
Parent[i] = Parent[4] = 1
So the condition is A[1] <= A[4]
i.e, 5 <= 15 so no need to swap
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
19

20. MIN-HEAP – DATA STRUCTURE
FIRST METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
As you can see that the min-heap is ready
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
20

21. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
In the second method, as we construct a complete binary
tree, we will check the condition i.e A[Parent[i]] < = A[i] for
every node that is inserted in the tree.
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
21

22. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
0
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
22

23. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 1
Parent[i] = Parent[1] = 0
Check the condition A[0]<=A[1]
15<=3 (false) so, we need to swap these two values
0
1
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
23

24. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 1
Parent[i] = Parent[1] = 0
Check the condition A[0]<=A[1]
3<=15 (true) so, continue
0
1
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
24

25. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 2
Parent[i] = Parent[2] = 0
Check the condition A[0]<=A[2]
i.e, 3<=15 so no need to swap
0
1 2
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
25

26. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 3
Parent[i] = Parent[3] = 1
Check the condition A[1]<=A[3]
i.e, 15<=18 so no need to swap
0
1 2
3
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
26

27. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 4
Parent[i] = Parent[4] = 1
Check the condition A[1]<=A[4]
i.e, 15<=5 (false) so we have to
swap
0
1 2
3 4
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
27

28. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
No the min-heap is okay
0
1 2
3 4
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
28

29. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 5
Parent[i] = Parent[5] = 2
Check the condition A[2]<=A[5]
i.e, 6<=9 so no need to swap
0
1 2
3 4 5
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
29

30. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
i = 6
Parent[i] = Parent[6] = 2
Check the condition A[2]<=A[6]
i.e, 6<=11 so no need to swap
0
1 2
3 4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
30

31. MIN-HEAP – DATA STRUCTURE
SECOND METHOD - 15, 3, 6, 18 ,5 ,9 ,11
Now Let’s check the condition for every node
Starting from leaf nodes
As you can see that the min-heap is ready
0
1
2
3
4 5 6
GOVERNMENT VIQAR-UN-NISA POST GRADUATE COLLEGE (W),
RAWALPINDI.
31