1. 1 CSE 326: Data Structures Part Four: Trees Henry Kautz Autumn 2002

2. 2 Material Weiss Chapter 4: N-ary trees Binary Search Trees AVL Trees Splay Trees

3. 3 Other Applications of Trees?

4. 4 Tree Jargon Length of a path = number of edges Depth of a node N = length of path from root to N Height of node N = length of longest path from N to a leaf Depth and height of tree = height of root A B CD E F depth=0, height = 2 depth = 2, height=0

5. 5 Definition and Tree Trivia Recursive Definition of a Tree: A tree is a set of nodes that is a. an empty set of nodes, or b. has one node called the root from which zero or more trees (subtrees) descend. A tree with N nodes always has ___ edges Two nodes in a tree have at most how many paths between them?

6. 6 Implementation of Trees Obvious Pointer-Based Implementation: Node with value and pointers to children –Problem? A B CD E F

7. 7 1 st Child/Next Sibling Representation Each node has 2 pointers: one to its first child and one to next sibling A B CD E F A B CD E F

8. 8 Nested List Implementation 1 Tree := ( label {Tree}* ) a d b c

9. 9 Nested List Implementation 2 Tree := label || (label {Tree}+ ) a d b c

10. 10 Example Arithmetic Expression: A + (B * (C / D) ) Tree for the above expression: Application: Arithmetic Expression Trees + A * B / C D Used in most compilers No parenthesis need – use tree structure Can speed up calculations e.g. replace / node with C/D if C and D are known Calculate by traversing tree (how?)

11. 11 Traversing Trees Preorder: Root, then Children –+ A * B / C D Postorder: Children, then Root –A B C D / * + Inorder: Left child, Root, Right child –A + B * C / D + A * B / C D

12. 12 void print_preorder ( TreeNode T) { TreeNode P; if ( T == NULL ) return; else { print_element(T.Element); P = T.FirstChild; while (P != NULL) { print_preorder ( P ); P = P.NextSibling; } } Example Code for Recursive Preorder What is the running time for a tree with N nodes?

13. 13 Binary Trees Properties Notation: depth(tree) = MAX {depth(leaf)} = height(root) –max # of leaves = 2 depth(tree) –max # of nodes = 2 depth(tree)+1 – 1 –max depth = n-1 –average depth for n nodes = (over all possible binary trees) Representation: A B DE C F HG JI Data right pointer left pointer

14. 14 Dictionary & Search ADTs Operations –create –destroy –insert –find –delete Dictionary: Stores values associated with user- specified keys –keys may be any (homogenous) comparable type –values may be any (homogenous) type –implementation: data field is a struct with two parts Search ADT: keys = values kim chi –spicy cabbage kreplach –tasty stuffed dough kiwi –Australian fruit insert find(kreplach) kohlrabi - upscale tuber kreplach - tasty stuffed dough

15. 15 Naïve Implementations unsorted array sorted array linked list insert (w/o duplicates) find delete Goal: fast find like sorted array, dynamic inserts/deletes like linked list

16. 16 Naïve Implementations unsorted array sorted array linked list insert (w/o duplicates) find + O(1)O(n)find + O(1) findO(n)O(log n)O(n) deletefind + O(1)O(n)find + O(1) Goal: fast find like sorted array, dynamic inserts/deletes like linked list

17. 17 Binary Search Tree Dictionary Data Structure 4 121062 115 8 14 13 79 Search tree property –all keys in left subtree smaller than root’s key –all keys in right subtree larger than root’s key –result: easy to find any given key inserts/deletes by changing links

18. 18 In Order Listing visit left subtree visit node visit right subtree 2092 155 10 307 17 In order listing:

19. 19 In Order Listing visit left subtree visit node visit right subtree 2092 155 10 307 17 In order listing: 2  5  7  9  10  15  17  20  30

20. 20 Finding a Node Node find(Comparable x, Node root) { if (root == NULL) return root; else if (x < root.key) return find(x,root.left); else if (x > root.key) return find(x, root.right); else return root; } 2092 155 10 307 17 runtime:

21. 21 Insert Concept: proceed down tree as in Find; if new key not found, then insert a new node at last spot traversed void insert(Comparable x, Node root) { // Does not work for empty tree – when root is NULL if (x < root.key){ if (root.left == NULL) root.left = new Node(x); else insert( x, root.left ); } else if (x > root.key){ if (root.right == NULL) root.right = new Node(x); else insert( x, root.right ); } }

22. 22 Time to Build a Tree Suppose a 1, a 2, …, a n are inserted into an initially empty BST: 1.a 1, a 2, …, a n are in increasing order 2.a 1, a 2, …, a n are in decreasing order 3.a 1 is the median of all, a 2 is the median of elements less than a 1, a 3 is the median of elements greater than a 1, etc. 4.data is randomly ordered

23. 23 Analysis of BuildTree Increasing / Decreasing:  (n 2 ) 1 + 2 + 3 + … + n =  (n 2 ) Medians – yields perfectly balanced tree  (n log n) Average case assuming all input sequences are equally likely is  (n log n) –equivalently: average depth of a node is log n Total time = sum of depths of nodes

24. 24 Proof that Average Depth of a Node in a BST Constructed from Random Data is  (log n) Method: Calculate sum of all depths, divide by number of nodes D(n) = sum of depths of all nodes in a random BST containing n nodes D(n) = D(left subtree)+D(right subtree) + adjustment for distance from root to subtree + depth of root D(n) = D(left subtree)+D(right subtree) + (number of nodes in left and right subtrees) + 0 D(n) = D(L)+D(n-L-1)+(n-1)

25. 25 Random BST, cont. D(n) = D(L)+D(n-L-1)+(n-1) For random data, all subtree sizes equally likely this looks just like the Quicksort average case equation!

26. 26

27. 27 Random Input vs. Random Trees Inputs 1,2,3 3,2,1 1,3,2 3,1,2 2,1,3 2,3,1 Trees For three items, the shallowest tree is twice as likely as any other – effect grows as n increases. For n=4, probability of getting a shallow tree > 50%

28. 28 Deletion 2092 155 10 307 17 Why might deletion be harder than insertion?

29. 29 FindMin/FindMax Node min(Node root) { if (root.left == NULL) return root; else return min(root.left); } 2092 155 10 307 17 How many children can the min of a node have?

30. 30 Successor Find the next larger node in this node’s subtree. –not next larger in entire tree Node succ(Node root) { if (root.right == NULL) return NULL; else return min(root.right); } 2092 155 10 307 17 How many children can the successor of a node have?

31. 31 Deletion - Leaf Case 2092 155 10 307 17 Delete(17)

32. 32 Deletion - One Child Case 2092 155 10 307 Delete(15)

33. 33 Deletion - Two Child Case 3092 205 10 7 Delete(5) replace node with value guaranteed to be between the left and right subtrees: the successor Could we have used the predecessor instead?

34. 34 Deletion - Two Child Case 3092 205 10 7 Delete(5) always easy to delete the successor – always has either 0 or 1 children!

35. 35 Deletion - Two Child Case 3092 207 10 7 Delete(5) Finally copy data value from deleted successor into original node

36. 36 Lazy Deletion Instead of physically deleting nodes, just mark them as deleted +simpler +physical deletions done in batches +some adds just flip deleted flag –extra memory for deleted flag –many lazy deletions slow finds –some operations may have to be modified (e.g., min and max) 2092 155 10 307 17

37. 37 Dictionary Implementations BST’s looking good for shallow trees, i.e. the depth D is small (log n), otherwise as bad as a linked list! unsorted array sorted array linked list BST insertfind + O(1) O(n)find + O(1) O(Depth) findO(n)O(log n)O(n)O(Depth) deletefind + O(1) O(n)find + O(1) O(Depth)

38. 38 CSE 326: Data Structures Part 3: Trees, continued Balancing Act Henry Kautz Autumn Quarter 2002

39. 39 Beauty is Only  (log n) Deep Binary Search Trees are fast if they’re shallow e.g.: complete Problems occur when one branch is much longer than the other How to capture the notion of a “sort of” complete tree?

40. 40 Balance balance = height(left subtree) - height(right subtree) convention: height of a “null” subtree is -1 zero everywhere  perfectly balanced small everywhere  balanced enough:  (log n) –Precisely: Maximum depth is 1.44 log n t 5 6

41. 41 AVL Tree Dictionary Data Structure 4 121062 115 8 141379 Binary search tree properties Balance of every node is -1  b  1 Tree re-balances itself after every insert or delete 15 What is the balance of each node in this tree?

42. 42 AVL Tree Data Structure 15 92 12 5 10 20 17 0 0 100 12 3 10 3 data height children 30 0

43. 43 Not An AVL Tree 15 92 12 5 10 20 17 0 1 200 13 4 10 4 data height children 30 0 18 0

44. 44 Bad Case #1 Insert(small) Insert(middle) Insert(tall) T M S 0 1 2

45. 45 Single Rotation T M S 0 1 2 M ST 00 1 Basic operation used in AVL trees: A right child could legally have its parent as its left child.

46. 46 General Case: Insert Unbalances a X Y b Z h h - 1 h + 1 h - 1 h + 2 a X Y b Z h-1 h h + 1 a XY b Z h h - 1 h h + 1

47. 47 Properties of General Insert + Single Rotation Restores balance to a lowest point in tree where imbalance occurs After rotation, height of the subtree (in the example, h+1) is the same as it was before the insert that imbalanced it Thus, no further rotations are needed anywhere in the tree!

48. 48 Bad Case #2 Insert(small) Insert(tall) Insert(middle) M T S 0 1 2 Why won’t a single rotation (bringing T up to the top) fix this?

49. 49 Double Rotation M ST 00 1 M T S 0 1 2 T M S 0 1 2

50. 50 General Double Rotation Initially: insert into X unbalances tree (root height goes to h+3) “Zig zag” to pull up c – restores root height to h+2, left subtree height to h a Z b W c Y a Z b W c Y h+1 h h h h + 3 h + 2 hh h+1 h + 2 h+1 h X X

51. 51 Another Double Rotation Case Initially: insert into Y unbalances tree (root height goes to h+2) “Zig zag” to pull up c – restores root height to h+1, left subtree height to h a Z b W c Y a Z b W c Y h+1 h h h h + 3 h + 2 h h h+1 h + 2 h+1 h X X

52. 52 Insert Algorithm Find spot for value Hang new node Search back up looking for imbalance If there is an imbalance: “outside”: Perform single rotation and exit “inside”: Perform double rotation and exit

53. 53 AVL Insert Algorithm Node insert(Comparable x, Node root){ // returns root of revised tree if ( root == NULL ) return new Node(x); if (x <= root.key){ root.left = insert( x, root.left ); if (root unbalanced) { rotate... } } else { // x > root.key root.right = insert( x, root.right ); if (root unbalanced) { rotate... } } root.height = max(root.left.height, root.right.height)+1; return root; }

54. 54 Deletion (Really Easy Case) 2092 155 10 30173 12 1 0 100 22 3 0 0 Delete(17)

55. 55 Deletion (Pretty Easy Case) 2092 155 10 30173 12 1 0 100 22 3 0 0 Delete(15)

56. 56 Deletion (Pretty Easy Case cont.) 2092 175 10 303 12 1 100 22 3 0 0 Delete(15)

57. 57 Deletion (Hard Case #1) 2092 175 10 303 12 1 100 22 3 0 0 Delete(12)

58. 58 Single Rotation on Deletion 2092 175 10 303 1 10 22 3 0 0 92 205 10 17 3 1 00 21 3 0 0 What is different about deletion than insertion?

59. 59 Deletion (Hard Case) Delete(9) 2092 175 10 303 12 1 220 23 4 0 33 15 13 00 1 0 20 30 12 33 15 13 1 00 11 0 18 0

60. 60 Double Rotation on Deletion 2 3 0 202 175 10 30 12 1 22 23 4 33 15 13 1 00 1 11 0 18 0 2052 173 10 30 12 0 220 13 4 33 15 13 1 00 1 11 0 18 00 Not finished!

61. 61 Deletion with Propagation We get to choose whether to single or double rotate! 2052 173 10 30 12 0 220 13 4 33 15 13 1 00 1 11 0 18 0 What different about this case?

62. 62 Propagated Single Rotation 0 30 20 17 33 12 15 13 1 0 52 3 10 4 32 121 000 11 0 2052 173 10 30 12 0 220 13 4 33 15 13 1 0 1 11 0 18 0 0

63. 63 Propagated Double Rotation 0 17 12 11 52 3 10 4 23 10 0 0 2052 173 10 30 12 0 220 13 4 33 15 13 1 0 1 11 0 18 0 15 1 0 20 30 33 1 18 0 13 0 2

64. 64 AVL Deletion Algorithm Recursive 1.If at node, delete it 2.Otherwise recurse to find it in 3. Correct heights a. If imbalance #1, single rotate b. If imbalance #2 (or don’t care), double rotate Iterative 1. Search downward for node, stacking parent nodes 2. Delete node 3. Unwind stack, correcting heights a. If imbalance #1, single rotate b. If imbalance #2 (or don’t care) double rotate

65. 65 AVL Automatically Virtually Leveled Architecture for inVisible Leveling Articulating Various Lines Amortizing? Very Lousy! Amazingly Vexing Letters

66. 66 AVL Automatically Virtually Leveled Architecture for inVisible Leveling Articulating Various Lines Amortizing? Very Lousy! Amazingly Vexing Letters Adelson-Velskii Landis

67. 67 Pro: All operations guaranteed O(log N) The height balancing adds no more than a constant factor to the speed of insertion Con: Space consumed by height field in each node Slower than ordinary BST on random data Can we guarantee O(log N) performance with less overhead? Pros and Cons of AVL Trees

68. 68 Splay Trees CSE 326: Data Structures Part 3: Trees, continued

69. 69 Today: Splay Trees Fast both in worst-case amortized analysis and in practice Are used in the kernel of NT for keep track of process information! Invented by Sleator and Tarjan (1985) Details: Weiss 4.5 (basic splay trees) 11.5 (amortized analysis) 12.1 (better “top down” implementation)

70. 70 Basic Idea “Blind” rebalancing – no height info kept! Worst-case time per operation is O(n) Worst-case amortized time is O(log n) Insert/find always rotates node to the root! Good locality: –Most commonly accessed keys move high in tree – become easier and easier to find

71. 71 Idea 17 10 92 5 3 You’re forced to make a really deep access: Since you’re down there anyway, fix up a lot of deep nodes! move n to root by series of zig-zag and zig-zig rotations, followed by a final single rotation (zig) if necessary

72. 72 Zig-Zag* g X p Y n Z W * This is just a double rotation n Y g W p ZX Helped Unchanged Hurt up 2 down 1 up 1down 1

73. 73 Zig-Zig n Z Y p X g W g W X p Y n Z

74. 74 Why Splaying Helps Node n and its children are always helped (raised) Except for last step, nodes that are hurt by a zig- zag or zig-zig are later helped by a rotation higher up the tree! Result: –shallow nodes may increase depth by one or two –helped nodes decrease depth by a large amount If a node n on the access path is at depth d before the splay, it’s at about depth d/2 after the splay –Exceptions are the root, the child of the root, and the node splayed

75. 75 Splaying Example 2 1 3 4 5 6 Find(6) 2 1 3 6 5 4 zig-zig

76. 76 Still Splaying 6 zig-zig 2 1 3 6 5 4 1 6 3 25 4

77. 77 Almost There, Stay on Target zig 1 6 3 25 4 6 1 3 25 4

78. 78 Splay Again Find(4) zig-zag 6 1 3 25 4 6 1 4 35 2

79. 79 Example Splayed Out zig-zag 6 1 4 35 2 61 4 35 2

80. 80 Locality “Locality” – if an item is accessed, it is likely to be accessed again soon –Why? Assume m  n access in a tree of size n –Total worst case time is O(m log n) –O(log n) per access amortized time Suppose only k distinct items are accessed in the m accesses. –Time is O(n log n + m log k ) –Compare with O( m log n ) for AVL tree getting those k items near root those k items are all at the top of the tree

81. 81 Splay Operations: Insert To insert, could do an ordinary BST insert –but would not fix up tree –A BST insert followed by a find (splay)? Better idea: do the splay before the insert! How?

82. 82 Split Split(T, x) creates two BST’s L and R: –All elements of T are in either L or R –All elements in L are  x –All elements in R are  x –L and R share no elements Then how do we do the insert?

83. 83 Split Split(T, x) creates two BST’s L and R: –All elements of T are in either L or R –All elements in L are  x –All elements in R are > x –L and R share no elements Then how do we do the insert? Insert as root, with children L and R

84. 84 Splitting in Splay Trees How can we split? –We have the splay operation –We can find x or the parent of where x would be if we were to insert it as an ordinary BST –We can splay x or the parent to the root –Then break one of the links from the root to a child

85. 85 Split split(x) TLR splay OR LRLR  x > x < x could be x, or what would have been the parent of x if root is  x if root is > x

86. 86 Back to Insert split(x) LR x LR > x  x Insert(x): Split on x Join subtrees using x as root

87. 87 Insert Example 91 6 47 2 Insert(5) split(5) 9 6 7 1 4 2 1 4 2 9 6 7 1 4 2 9 6 7 5

88. 88 Splay Operations: Delete find(x) LR x LR > x< x delete x Now what?

89. 89 Join Join(L, R): given two trees such that L < R, merge them Splay on the maximum element in L then attach R LR R splay L

90. 90 Delete Completed T find(x) LR x LR > x< x delete x T - x Join(L,R)

91. 91 Delete Example 91 6 47 2 Delete(4) find(4) 9 6 7 1 4 2 1 2 9 6 7 Find max 2 1 9 6 7 2 1 9 6 7

92. 92 Splay Trees, Summary Splay trees are arguably the most practical kind of self-balancing trees If number of finds is much larger than n, then locality is crucial! –Example: word-counting Also supports efficient Split and Join operations – useful for other tasks –E.g., range queries