Programming Binary Search Tree. Virendra Singh Indian Institute of Science Bangalore Lecture 12

Transcription

1 SE-286: Data Structures and Programming g Binary Search Tree Virendra Singh Indian Institute of Science Bangalore Lecture 12 Courtesy: Prof. Sartaj Sahni 1

2 Binary Search Trees Dictionary Operations: get(key) put(key, value) remove(key) Additional operations: ascend() get(index) (indexed binary search tree) remove(index) (indexed binary search tree) Sep 8, 21 2

3 Complexity Of Dictionary Operations get(), put() and remove() Data Structure Worst Case Expected Hash htable O(n) O(1) Binary Search O(n) O(log n) Tree Balanced O(log n) O(log n) Binary Search Tree Sep n8, 21 is number of elements dictionary 3

4 Complexity Of Other Operations ascend(), get(index), remove(index) Data Structure ascend get and remove Hash Table O(D + n log n) O(D + n log n) Indexed BST O(n) O(n) Indexed Balanced BST O(n) O(log n) Sep D8, 21 is number of buckets SE286@SERC 4

5 Definition Of Binary Search Tree A binary tree. Each node has a (key, value) pair. For every node x, all keys in the left subtree of x are smaller than that in x. For every node x, all keys in the right subtree of x are greater than that in x. Sep 8, 21 SE286@SERC 5

6 Example Binary Search Tree Only keys are shown. Sep 8, 21 SE286@SERC 6

7 The Operation ascend() Do an inorder traversal. O(n) time. Sep 8, 21 SE286@SERC

8 The Operation get() Complexity is O(height) = O(n), where n is number of nodes/elements. Sep 8, 21 SE286@SERC 8

9 The Operation put() Put Sep a 8, 21 pair whose key is SE286@SERC 35. 9

10 The Operation put() Put Sep a 8, 21 pair whose key is SE286@SERC. 1

11 The Operation put() Put a pair whose key is 18. Sep 8, 21 SE286@SERC 11

12 The Operation put() Complexity Sep 8, 21 of put() is SE286@SERC O(height). 12

13 The Operation remove() Three cases: Element is in a leaf. Element is in a degree 1 node. Element is in a degree 2 node. Sep 8, 21 SE286@SERC 13

14 Remove From A Leaf Remove a leaf element. key = Sep 8, 21 SE286@SERC 14

15 Remove From A Leaf (contd.) Remove a leaf element. key = 35 Sep 8, 21 SE286@SERC 15

16 Remove From A Degree 1 Node Remove from a degree 1 node. key = 4 Sep 8, 21 SE286@SERC 16

17 Remove From A Degree 1 Node (contd.) Remove from a degree 1 node. key = 15 Sep 8, 21 SE286@SERC 1

18 Remove From A Degree 2 Node Remove from a degree 2 node. key = 1 Sep 8, 21 SE286@SERC 18

19 Remove From A Degree 2 Node Replace with largest key in left subtree (or smallest in right subtree). 19

20 Remove From A Degree 2 Node Replace with largest key in left subtree (or smallest in right subtree). 2

21 Remove From A Degree 2 Node Replace with largest key in left subtree (or smallest in right subtree). Sep 8, 21 SE286@SERC 21

22 Remove From A Degree 2 Node Largest key must be in a leaf or degree 1 node. Sep 8, 21 SE286@SERC 22

23 Another Remove From A Degree 2 Node Remove from a degree 2 node. key = 2 Sep 8, 21 SE286@SERC 23

24 Remove From A Degree 2 Node Replace with largest in left subtree. Sep 8, 21 SE286@SERC 24

25 Remove From A Degree 2 Node Replace with largest in left subtree. Sep 8, 21 SE286@SERC 25

26 Remove From A Degree 2 Node Replace with largest in left subtree. Sep 8, 21 SE286@SERC 26

27 Remove From A Degree 2 Node Complexity is O(height). Sep 8, 21 SE286@SERC 2

28 Indexed Binary Search Tree Binary search tree. Each node has an additional field. leftsize = number of nodes in its left subtree Sep 8, 21 SE286@SERC 28

29 Example Indexed Binary Search Tree leftsize values are in red Sep 8, 21 SE286@SERC 29

30 leftsize And Rank Rank of an element is its position in inorder (inorder = ascending key order). rank(2) = rank(15) = 5 rank(2) = [2,6,,8,1,15,18,2,25,3,35,4] leftsize(x) = rank(x) with respect to elements in subtree rooted at x Sep 8, 21 SE286@SERC 3

31 1 leftsize And Rank sorted list = [2,6,,8,1,15,18,2,25,3,35,4] Sep 8, 21 SE286@SERC 31

32 get(index) And remove(index) sorted list = [2,6,,8,1,15,18,2,25,3,35,4] Sep 8, 21 SE286@SERC 32

33 get(index) And remove(index) if index = x.leftsize desired element is x.element if index < x.leftsize desired element is index th element in left subtree of x if index > x.leftsize desired element is (index - x.leftsize-1) 1) th element in right subtree of x Sep 8, 21 SE286@SERC 33

34 Applications (Complexities Are For Balanced Trees) Best-fit bin packing in O(n log n) time. Representing a linear list so that tget(index), add(index, element), and remove(index) run in O(log(list (li size)) time (uses an indexed d binary tree, not indexed binary search tree). Can t use hash tables for either of these applications. Sep 8, 21 SE286@SERC 34

35 Linear List As Indexed Binary Tree 1 e h 4 3 b f j 1 1 g a d i k c list = [a,b,c,d,e,f,g,h,i,j,k,l] ijk Sep 8, 21 SE286@SERC 35 l

36 add(5, m ) 1 4 e b f j 1 h 1 g a d i k c l 3 list =[abcdefghijkl] [a,b,c,d,e,f,g,h,i,j,k,l] Sep 8, 21 SE286@SERC 36

37 add(5, m ) 1 4 e b f j 1 h 1 g a d i k c l 3 list =[abcde [a,b,c,d,e, m,f,g,h,i,j,k,l] fghijkl] find node with element 4 (e) 3

38 add(5, m ) 1 4 e b f j 1 h 1 g a d i k c l 3 list =[abcde [a,b,c,d,e, m,f,g,h,i,j,k,l] fghijkl] find node with element 4 (e) 38

39 add(5, m ) 1 4 e m b f j 1 h 1 g a d i k c l 3 add m as right child of e; former right subtree of e becomes right subtree of m 39

40 add(5, m ) 1 4 e b f j 1 h 1 a d m g i k c l 3 add m as lf leftmost node in right ihsubtree of e Sep 8, 21 SE286@SERC 4

41 add(5, m ) Other possibilities exist. Must update some leftsize values on path from root to new node. Complexity is O(height). Sep 8, 21 SE286@SERC 41