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CS 332: Algorithms

CS 332: Algorithms. Data Structures For Dynamic Sets Binary Search Trees. Dynamic Sets. Next few lectures will focus on data structures rather than straight algorithms In particular, structures for dynamic sets Elements have a key and satellite data

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CS 332: Algorithms

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  1. CS 332: Algorithms Data Structures For Dynamic SetsBinary Search Trees David Luebke 110/2/2014

  2. Dynamic Sets • Next few lectures will focus on data structures rather than straight algorithms • In particular, structures for dynamic sets • Elements have a key and satellite data • Dynamic sets support queries such as: • Search(S, k), Minimum(S), Maximum(S), Successor(S, x), Predecessor(S, x) • They may also support modifying operationslike: • Insert(S, x), Delete(S, x) David Luebke 210/2/2014

  3. Binary Search Trees • Binary Search Trees (BSTs) are an important data structure for dynamic sets • In addition to satellite data, eleements have: • key: an identifying field inducing a total ordering • left: pointer to a left child (may be NULL) • right: pointer to a right child (may be NULL) • p: pointer to a parent node (NULL for root) David Luebke 310/2/2014

  4. F B H A D K Binary Search Trees • BST property: key[left(x)]  key[x]  key[right(x)] • Example: David Luebke 410/2/2014

  5. Inorder Tree Walk • What does the following code do? TreeWalk(x) TreeWalk(left[x]); print(x); TreeWalk(right[x]); • A: prints elements in sorted (increasing) order • This is called an inorder tree walk • Preorder tree walk: print root, then left, then right • Postorder tree walk: print left, then right, then root David Luebke 510/2/2014

  6. F B H A D K Inorder Tree Walk • Example: • How long will a tree walk take? • Prove that inorder walk prints in monotonically increasing order David Luebke 610/2/2014

  7. Operations on BSTs: Search • Given a key and a pointer to a node, returns an element with that key or NULL: TreeSearch(x, k) if (x = NULL or k = key[x]) return x; if (k < key[x]) return TreeSearch(left[x], k); else return TreeSearch(right[x], k); David Luebke 710/2/2014

  8. F B H A D K BST Search: Example • Search for D and C: David Luebke 810/2/2014

  9. Operations on BSTs: Search • Here’s another function that does the same: TreeSearch(x, k) while (x != NULL and k != key[x]) if (k < key[x]) x = left[x]; else x = right[x]; return x; • Which of these two functions is more efficient? David Luebke 910/2/2014

  10. Operations of BSTs: Insert • Adds an element x to the tree so that the binary search tree property continues to hold • The basic algorithm • Like the search procedure above • Insert x in place of NULL • Use a “trailing pointer” to keep track of where you came from (like inserting into singly linked list) David Luebke 1010/2/2014

  11. F B H A D K BST Insert: Example • Example: Insert C C David Luebke 1110/2/2014

  12. BST Search/Insert: Running Time • What is the running time of TreeSearch() or TreeInsert()? • A: O(h), where h = height of tree • What is the height of a binary search tree? • A: worst case: h = O(n) when tree is just a linear string of left or right children • We’ll keep all analysis in terms of h for now • Later we’ll see how to maintain h = O(lg n) David Luebke 1210/2/2014

  13. Sorting With Binary Search Trees • Informal code for sorting array A of length n: BSTSort(A) for i=1 to n TreeInsert(A[i]); InorderTreeWalk(root); • Argue that this is (n lg n) • What will be the running time in the • Worst case? • Average case? (hint: remind you of anything?) David Luebke 1310/2/2014

  14. 3 1 8 2 6 5 7 Sorting With BSTs • Average case analysis • It’s a form of quicksort! for i=1 to n TreeInsert(A[i]); InorderTreeWalk(root); 3 1 8 2 6 7 5 1 2 8 6 7 5 2 6 7 5 5 7 David Luebke 1410/2/2014

  15. Sorting with BSTs • Same partitions are done as with quicksort, but in a different order • In previous example • Everything was compared to 3 once • Then those items < 3 were compared to 1 once • Etc. • Same comparisons as quicksort, different order! • Example: consider inserting 5 David Luebke 1510/2/2014

  16. Sorting with BSTs • Since run time is proportional to the number of comparisons, same time as quicksort: O(n lg n) • Which do you think is better, quicksort or BSTsort? Why? David Luebke 1610/2/2014

  17. Sorting with BSTs • Since run time is proportional to the number of comparisons, same time as quicksort: O(n lg n) • Which do you think is better, quicksort or BSTSort? Why? • A: quicksort • Better constants • Sorts in place • Doesn’t need to build data structure David Luebke 1710/2/2014

  18. More BST Operations • BSTs are good for more than sorting. For example, can implement a priority queue • What operations must a priority queue have? • Insert • Minimum • Extract-Min David Luebke 1810/2/2014

  19. BST Operations: Minimum • How can we implement a Minimum() query? • What is the running time? David Luebke 1910/2/2014

  20. BST Operations: Successor • For deletion, we will need a Successor() operation • Draw Fig 13.2 • What is the successor of node 3? Node 15? Node 13? • What are the general rules for finding the successor of node x? (hint: two cases) David Luebke 2010/2/2014

  21. BST Operations: Successor • Two cases: • x has a right subtree: successor is minimum node in right subtree • x has no right subtree: successor is first ancestor of x whose left child is also ancestor of x • Intuition: As long as you move to the left up the tree, you’re visiting smaller nodes. • Predecessor: similar algorithm David Luebke 2110/2/2014

  22. F Example: delete Kor H or B B H C A D K BST Operations: Delete • Deletion is a bit tricky • 3 cases: • x has no children: • Remove x • x has one child: • Splice out x • x has two children: • Swap x with successor • Perform case 1 or 2 to delete it David Luebke 2210/2/2014

  23. BST Operations: Delete • Why will case 2 always go to case 0 or case 1? • A: because when x has 2 children, its successor is the minimum in its right subtree • Could we swap x with predecessor instead of successor? • A: yes. Would it be a good idea? • A: might be good to alternate David Luebke 2310/2/2014

  24. The End • Up next: guaranteeing a O(lg n) height tree David Luebke 2410/2/2014

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