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Initializing A Max Heap

1. 3. 2. 4. 5. 6. 7. 11. 8. 9. 7. 7. 8. 10. Initializing A Max Heap. input array = [-, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Initializing A Max Heap. 1. 3. 2. 4. 5. 6. 7. Start at rightmost array position that has a child. 11. 8. 9. 7. 7. 8. 10. Index is n/2.

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Initializing A Max Heap

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  1. 1 3 2 4 5 6 7 11 8 9 7 7 8 10 Initializing A Max Heap input array = [-, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

  2. Initializing A Max Heap 1 3 2 4 5 6 7 Start at rightmost array position that has a child. 11 8 9 7 7 8 10 Index is n/2.

  3. Initializing A Max Heap 1 3 2 4 11 6 7 Move to next lower array position. 5 8 9 7 7 8 10

  4. Initializing A Max Heap 1 3 2 4 11 6 7 5 8 9 7 7 8 10

  5. Initializing A Max Heap 1 3 2 9 11 6 7 5 8 4 7 7 8 10

  6. Initializing A Max Heap 1 3 2 9 11 6 7 5 8 4 7 7 8 10

  7. Initializing A Max Heap 1 7 2 9 11 6 3 5 8 4 7 7 8 10

  8. Initializing A Max Heap 1 7 2 9 11 6 3 5 8 4 7 7 8 10

  9. Initializing A Max Heap 1 7 11 9 6 3 5 8 4 7 7 8 10 Find a home for 2.

  10. Initializing A Max Heap 1 7 11 9 10 6 3 8 4 8 7 7 5 Find a home for 2.

  11. Initializing A Max Heap 1 7 11 9 10 6 3 8 4 7 7 8 5 2 Done, move to next lower array position.

  12. Initializing A Max Heap 1 7 11 9 10 6 3 2 8 4 7 7 8 5 Find home for 1.

  13. Initializing A Max Heap 11 7 9 10 6 3 2 8 4 7 7 8 5 Find home for 1.

  14. Initializing A Max Heap 11 10 7 9 6 3 2 8 4 7 7 8 5 Find home for 1.

  15. Initializing A Max Heap 11 10 7 9 5 6 3 2 8 4 7 7 8 Find home for 1.

  16. Initializing A Max Heap 11 10 7 9 5 6 3 2 8 4 7 7 8 1 Done.

  17. Time Complexity 11 7 9 8 5 6 3 1 4 7 7 8 2 10 Height of heap = h. Number of subtrees with root at level j is <= 2 j-1. Time for each subtree is O(h-j+1).

  18. Complexity Time for level j subtrees is <= 2j-1(h-j+1) = t(j). Total time is t(1) + t(2) + … + t(h-1) =

  19. Heap Sort • Heap Sort (program 9.12) • create a max heap • keep extracting the root element from the max heap, and put the element into the result array accordingly • time complexity = initialization time + n × deleting element time = O(n)+O(n × log n) = O(n log n)

  20. Heap Sort --example

  21. Comparison of sorting methods

  22. Comparison of sorting methods

  23. Comparison of sorting methods

  24. Leftist Tree Linked binary tree. Can do everything a heap can do and in the same asymptotic complexity. Can meld two leftist tree priority queues in O(log n) time.

  25. Extended Binary Trees Start with any binary tree and add an external node wherever there is an empty subtree. Result is an extended binary tree.

  26. A Binary Tree

  27. An Extended Binary Tree number of external nodes is n+1

  28. The Function s() For any node x in an extended binary tree, let s(x) be the length of a shortest path from x to an external node in the subtree rooted at x.

  29. s() Values Example

  30. s() Values Example 2 2 1 2 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0

  31. Properties Of s() If x is an external node, then s(x) = 0. Otherwise, s(x) = min {s(leftChild(x)), s(rightChild(x))} + 1

  32. Height Biased Leftist Trees A binary tree is a (height biased) leftist tree iff for every internal node x, s(leftChild(x)) >= s(rightChild(x))

  33. A Leftist Tree 2 2 1 2 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0

  34. Leftist Trees--Property 1 In a leftist tree, the rightmost path is a shortest root to external node path and the length of this path is s(root).

  35. A Leftist Tree 2 2 1 2 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 Length of rightmost path is 2.

  36. Leftist Trees—Property 2 The number of internal nodes is at least 2s(root) - 1 (note: because level 1 through level s(root) have no external nodes.) If there are n nodes in the leftist tree, then s(root)  log (n+1) (because 2s(root) – 1  n )

  37. A Leftist Tree 2 2 1 2 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 Levels 1 and 2 have no external nodes.

  38. Leftist Trees—Property 3 Length of rightmost path is O(log n), where n is the number of nodes in a leftist tree. Follows from Properties 1 and 2.

  39. Leftist Trees As Priority Queues • Min leftist tree … leftist tree that is a min tree. • Used as a min priority queue. • Max leftist tree … leftist tree that is a max tree. • Used as a max priority queue.

  40. A Min Leftist Tree 2 4 3 6 8 5 9 8 6

  41. Some Min Leftist Tree Operations put() remove() meld() initialize() put() and remove() use meld().

  42. Put Operation 2 4 3 put(7) 6 8 5 9 8 6

  43. Meld operation • The meld strategy is the best described using recursion • Let A and B be the two min HBLTs that are to be melded • The root with smaller element becomes the root of the melded HBLT

  44. Meld operation – cont. • Suppose that A has the smaller root and that its left subtree is L. • Let C be the min HBLT that results from melding the right subtree of A and the min HBLT B • The result of melding A and B is the min HBLT has A as its root and L and C as its subtrees. • If the s values of L is smaller than that of C, the C is the left subtree. Otherwise, L is.

  45. Put Operation 2 4 3 put(7) 6 8 5 9 8 6 Create a single node min leftist tree. 7

  46. Put Operation 2 4 3 put(7) 6 8 5 9 8 6 Create a single node min leftist tree. Meld the two min leftist trees. 7

  47. Remove Min 2 4 3 6 8 5 9 8 6

  48. Remove Min 2 4 3 6 8 5 9 8 6 Remove the root.

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