Binary Search Trees Readings: Chapter 11

Transcription

1 Binary Search Trees Readings: Chapter 11 CSE214: Computer Science II Fall 2016 CHEN-WEI WANG

2 Binary Search Tree: Definition A binary search tree ( BST ) is a binary tree, where each internal node p stores a key-value pair (k, v), such that: For each node n of the LST of p: key(n) < key(p) For each node n of the RST of p: key(n) > key(p) Node p stores (key(p), value(p)) Each node n of LST is such that key(n) < key(p) Each node n of RST is such that key(n) > key(p) A BST can be used to efficiently implement a sorted map. ) All stored keys are unique (i.e., no duplicates). 2 of 39

3 BST: Sorting Property An in-order traversal of the keys stored in a BST will result in a sequence arranged in an increasing order Justification: In-order traversal: visit LST, then root, then RST 3 of 39 BST : keys in LST < root s key and keys in RST > root s key 80

4 BST: Internal Nodes vs. External Nodes We store key-value pairs only in internal nodes. We treat external nodes as sentinels, in order to simplify the definitions of BST algorithms. Similar to how we treat header and trailer in a doubly-linked list 4 of 39

5 BST Operation: Searching (1) We may attempt to locate a particular key k in a BST rooted at node p, by viewing it as a decision tree. BinarySearch(p, k): if(isexternal(p)) { return p; /* unsuccessful search */ } else if (k == key(p)) then { return p; /* successful search */ } else if (k < key(p)) then { return BinarySearch(left(p), k); /* recur on LST */ } else { /* k > key(p) */ return BinarySearch(right(p), k); /* recur on RST */ } 5 of 39

6 BST Operation: Searching (2) A successful search for key of 39

7 BST Operation: Searching (3) An unsuccessful search for key The search terminates at left external child of the key 76 7 of 39

8 BST Operation: RT of Searching (1) Recursive calls of BinarySearch are made on nodes of path which Starts from the root Goes down one level at a time Stops when the key is found, or when a leaf is reached Maximum number of nodes visited by the search? [h + 1] RT of deciding from each node to go to LST or RST? [O(1)] So RT of a binary search is O(h) Recall: Given a binary tree with n modes, the height h is bounded as log(n + 1) 1 apple h apple n 1 Best RT of a binary search is O(log(n)) Worst RT of a binary search is O(n) 8 of 39 [balanced BT] [ill-balanced BT]

9 BST Operation: RT of Searching (2) Height Time per level O(1) Tree T: O(1) h O(1) Total time: O(h) 9 of 39

10 BST Operation: Insertion (1) We may insert an entry with key k and value v into a BST. We first need to know if k is an existing key in the BST. BSTInsert(k, v): p = BinarySearch(root(), k) if (k == key(p)) { /* k is an existing key */ set p s value to v } else { /* k is not an existing key */ turn p into an internal node storing (k, v) } Running time? [ O(h) ] 10 of 39

11 BST Operation: Insertion (2.1) Insert an entry with key 68 into the following BST: of 39

12 BST Operation: Insertion (2.2) After inserting an entry with key 68 into the above BST: of 39

13 BST Operation: Deletion (1) We may delete a node storing the key k from a BST. We first call p = BinarySearch(root(), k) to check if k is an existing key in the BST. Case 1: Node p is external k is not an existing key ) Nothing to remove Case 2: Node p s both children are external No orphan subtrees to be handled ) Remove p [BST property?] Case 3: One of the node p s children, say r, is internal r s sibling is external ) Replace p with r [BST property?] Case 4: Both of node p s children are internal Locate r containing the greatest key s.t. key(r) < key(p) ) r is the right-most internal node r of the LST of p Replace p s entry with r s entry [BST property?] r being the right-most internal node does not have an internal right child ) Remove r as defined in Case 3 [BST Property?] Running time? [ O(h) ] 13 of 39

14 BST Operation: Deletion (2.1) Delete the node storing key 32 from the following BST: p 88 8 r of 39

15 BST Operation: Deletion (2.2) After deleting the node storing key 32 from the above BST: r of 39

16 BST Operation: Deletion (3.1) Delete the node storing key 88 from the following BST: p r of 39

17 BST Operation: Deletion (3.2) After deleting the node storing key 88 from the above BST: p r of 39

18 Homework (1) Study the Java implementation of a sorted Map using a binary search tree (Section ). 18 of 39

19 Balanced Binary Search Trees: Motivation After insertions into a BST, the worst-case RT of a search occurs when the height h is at its maximum: O(n) : Entries were inserted in an decreasing order of their keys ) One-path, left-slanted BST [e.g., h100, 75, 68, 60, 50, 1i] Entries were inserted in an increasing order of their keys ) One-path, right-slanted BST [e.g., h1, 50, 60, 68, 75, 100i] Last entry s key is in-between keys of the previous two entries ) One-path, side-alternating BST [e.g., h1, 100, 50, 75, 60, 68i] To avoid the worst-case RT (* a ill-balanced tree), we need to take actions as soon as the tree becomes unbalanced. 19 of 39

20 Balanced Binary Search Trees: Definition Given a position (or a node) p, the height of the subtree rooted at p is recursively defined as: ( 0 if p is external height (p) = 1 + MAX ({ height (c) parent (c) =p }) if p is internal A balanced BST T satisfies the height-balance property : For every internal node v, heights of v s child nodes differ apple of 39 Q: Will the tree remain balanced after inserting 55? [NO] Q: Will the tree remain balanced after inserting 63? [YES ]

21 Fixing Unbalanced BST: Rotations A tree rotation is performed: When the latest insertion/deletion creates unbalanced nodes, along the ancestor path of the node being inserted/deleted. To change the shape of tree, maintaining the search property h + 2 h + 2 a b h + 1 h h b c rotate node b a h + 1 h c h T4 to the right h h T3 T1 T2 T3 T4 T1 T2 An in-order traversal on the resulting tree still produces a sorted sequence of keys. [ht 1, c, T 2, b, T 3, a, T 4 i] After rotating node b to the right: Heights of descendants (c, T 1, T 2, T 3 ) and sibling (T 4 ) remain same. Height of parent (a) is decreased by 1. ) If node a was unbalanced, then this rotation may have fixed it. 21 of 39

22 After Insertions: Trinode Restructuring via Rotation(s) After inserting a new node v: Case 1: Nodes alone the ancestor path of v remains balanced ) No rotations needed Case 2: Found a non-empty sequence of nodes alone the ancestor path of v that become unbalanced 1. Get the first (i.e., lowest) unbalanced node a in v s ancestor path 2. Get a s child node b in v s ancestor path 3. Get b s child node c in v s ancestor path 4. Perform rotation(s) based on the alignment of a, b, and c: Slanted the same way ) single rotation on the middle node b Slanted different ways ) double rotations on the lower node c 22 of 39

23 Trinode Restructuring: Single, Left Rotation a b T1 c T2 T3 T4 Left Rotation on node b (height of unbalanced subtree #) b a c T1 T2 T3 T4 BST property maintained? [ht 1, a, T 2, b, T 3, c, T 4 i] 23 of 39

24 Exercise: Left Rotation Insert the following sequence of nodes into an empty BST: h44, 17, 78, 32, 50, 88, 95i Is the BST now balanced? Insert 100 into the BST Is the BST now balanced? Perform the left rotation on the appropriate node. 24 of 39

25 Trinode Restructuring: Single, Right Rotation a b c T4 T3 T1 T2 Right Rotation on node b (height of unbalanced subtree #) b c a T1 T2 T3 T4 25 of 39 BST property maintained? [ht 1, a, T 2, b, T 3, c, T 4 i]

26 Exercise: Right Rotation Insert the following sequence of nodes into an empty BST: h44, 17, 78, 32, 50, 88, 48i Is the BST now balanced? Insert 46 into the BST Is the BST now balanced? Perform the right rotation on the appropriate node. 26 of 39

27 Trinode Restructuring: Double, R-L Rotations a b T1 c T4 T2 T3 Right Rotation on node c a c T1 b T2 T3 T4 Then Left Rotation on c (height of unbalanced subtree #) 27 of 39

28 Exercise: R-L Rotations Insert the following sequence of nodes into an empty BST: h44, 17, 78, 32, 50, 88, 82, 95i Is the BST now balanced? Insert 85 into the BST Is the BST now balanced? Perform the R-L rotations on the appropriate node. 28 of 39

29 Trinode Restructuring: Double, L-R Rotations a b c T4 T1 T2 T3 Left Rotation on node c a c b T4 T3 T1 T2 Then Right Rotation on c (height of unbalanced subtree #) 29 of 39

30 Exercise: L-R Rotations Insert the following sequence of nodes into an empty BST: h44, 17, 78, 32, 50, 88, 48, 62i Is the BST now balanced? Insert 54 into the BST Is the BST now balanced? Perform the L-R rotations on the appropriate node. 30 of 39

31 After Deletions: Continuous Trinode Restructuring After deleting an existing node, whose only child is v: Case 1: Nodes alone the ancestor path of v remains balanced ) No rotations needed Case 2: Found a non-empty sequence of nodes alone the ancestor path of v that become unbalanced 1. Get the first (i.e., lowest) unbalanced node a in v s ancestor path 2. Get a s taller child node b [b 62 v s ancestor path] 3. Choose b s child node c as follows: b s two child nodes have different heights ) c is the taller child b s two child nodes have same height ) a, b, c slant the same way 4. Perform rotation(s) based on the alignment of a, b, and c: Slanted the same way ) single rotation on the middle node b Slanted different ways ) double rotations on the lower node c Continue to apply Case 2 (i.e., unbalanced node found along v s ancestor path) until Case 1 is satisfied. [ O(log n) rotations] 31 of 39

32 Exercise: Single Trinode Restructuring Step Insert the following sequence of nodes into an empty BST: h44, 17, 62, 32, 50, 78, 48, 54, 88i Is the BST now balanced? Delete 32 from the BST Is the BST now balanced? Perform the left rotation on the appropriate node. Is the BST now balanced? 32 of 39

33 Exercise: Multiple Trinode Restructuring Steps Insert the following sequence of nodes into an empty BST: h50, 25, 10, 30, 5, 15, 27, 1, 75, 60, 80, 55i Is the BST now balanced? Delete 80 from the BST Is the BST now balanced? Perform the right rotation on the appropriate node. Is the BST now balanced? Perform the right rotation on the appropriate node. Is the BST now balanced? 33 of 39

34 Restoring Balance from Insertions Before Insertion into T3 After Insertion into T3 h + 3 h + 4 a a h + 2 b h + 1 h + 3 b h + 1 h + 1 h + 1 c T4 h + 1 h + 2 c T4 T1 h h T1 h h + 1 T2 T3 T2 T3 After Performing L-R Rotations on Node c: Height of Subtree Being Fixed Remains h + 3 h + 3 c h + 2 b h + 2 a h + 1 h h + 1 h + 1 T1 T2 T3 T4 34 of 39

35 Restoring Balance from Deletions Before Deletion from T4 After Deletion from T4 h + 3 a h + 3 a h + 2 b h + 1 h + 2 b h h + 1 c h T4 h + 1 c h T4 h h T3 h h T3 T1 T2 T1 T2 After Performing Right Rotation on Node b: Height of Subtree Being Fixed Reduces its Height by 1! h + 2 b h + 1 c h + 1 a h h h h T1 T2 T3 T4 35 of 39

36 Restoring Balance: Insertions vs. Deletions Each rotation involves only primitive assignments to re-assign parent-child references ) O(1) running time for each tree rotation After each insertion, a trinode restructuring step can restore the height of the subtree rooted at the first-encountered unbalanced node. ) O(1) rotations suffices to restore the balance of tree After each deletion, a trinode restructuring step may reduce the height of the subtree rooted at the first-encountered unbalanced node. ) May take O(log n) rotations to restore the balance of tree 36 of 39

37 Index (1) Binary Search Tree: Definition BST: Sorting Property BST: Internal Nodes vs. External Nodes BST Operation: Searching (1) BST Operation: Searching (2) BST Operation: Searching (3) BST Operation: RT of Searching (1) BST Operation: RT of Searching (2) BST Operation: Insertion (1) BST Operation: Insertion (2.1) BST Operation: Insertion (2.2) BST Operation: Deletion (1) BST Operation: Deletion (2.1) BST Operation: Deletion (2.2) 37 of 39

38 Index (2) BST Operation: Deletion (3.1) BST Operation: Deletion (3.2) Homework (1) Balanced Binary Search Trees: Motivation Balanced Binary Search Trees: Definition Fixing Unbalanced BST: Rotations After Insertions: Trinode Restructuring via Rotation(s) Trinode Restructuring: Single, Left Rotation Exercise: Left Rotation Trinode Restructuring: Single, Right Rotation Exercise: Right Rotation Trinode Restructuring: Double, R-L Rotations Exercise: R-L Rotations 38 of 39

39 Index (3) Trinode Restructuring: Double, L-R Rotations Exercise: L-R Rotations After Deletions: Continuous Trinode Restructuring Exercise: Single Trinode Restructuring Step Exercise: Multiple Trinode Restructuring Steps Restoring Balance from Insertions Restoring Balance from Deletions Restoring Balance: Insertions vs. Deletions 39 of 39