CSE 332: Data Structures & Parallelism Lecture 8: AVL Trees. Ruth Anderson Winter 2019

Transcription

1 CSE 2: Data Structures & Parallelism Lecture 8: AVL Trees Rut Anderson Winter 29

2 Today Dictionaries AVL Trees /25/29 2

3 Te AVL Balance Condition: Left and rigt subtrees of every node ave eigts differing by at most Define: balance(x) = eigt(x.left) eigt(x.rigt) AVL property: balance(x), for every node x Ensures small dept Will prove tis by sowing tat an AVL tree of eigt must ave a lot of (*rougly* 2 ) nodes Easy to maintain Using single and double rotations Note: eigt of a null tree is -, eigt of tree wit a single node is /25/29

4 Te AVL Tree Data Structure Structural properties. Binary tree property (,, or 2 cildren) 8 2. Heigts of left and rigt subtrees of every node differ by at most 5 Result: Worst case dept of any node 2 6 is: O(log n) Ordering property Same as for BST 5 /25/29 4

5 Ex: An AVL tree? /25/29 5

6 Ex2: An AVL tree? /25/29 7

7 Ex: An AVL tree? /25/29 9

8 Heigt of an AVL Tree? Using te AVL balance property, we can determine te minimum number of nodes in an AVL tree of eigt Let S()be te minimum # of nodes in an AVL tree of eigt, ten: S() = S(-) + S(-2) + were S(-) = and S() = Solution of Recurrence: S().62 /25/29

9 Let S()be te minimum # of nodes in an AVL tree of eigt, ten: S() = S(-) + S(-2) + were S(-)= and S()= Min eigt Tree S() /25/29 2

10 Minimal AVL Tree (eigt = ) /25/29

11 Minimal AVL Tree (eigt = ) /25/29 4

12 Minimal AVL Tree (eigt = 2) /25/29 5

13 Minimal AVL Tree (eigt = ) /25/29 6

14 Minimal AVL Tree (eigt = 4) /25/29 7

15 Te sallowness bound Let S() = te minimum number of nodes in an AVL tree of eigt If we can prove tat S() grows exponentially in, ten a tree wit n nodes as a logaritmic eigt Step : Define S() inductively using AVL property S(-)=, S()=, S()=2 For, S() = +S(-)+S(-2) -2 Step 2: Sow tis recurrence grows really fast Similar to Fibonacci numbers Can prove for all, S() > were is te golden ratio, (+ 5)/2, about.62 Growing faster tan.6 is plenty exponential - /25/29 8

16 Before we prove it Good intuition from plots comparing: S() computed directly from te definition ((+ 5)/2) S() is always bigger, up to trees wit uge numbers of nodes Graps aren t proofs, so let s prove it /25/29 9

17 Te Golden Ratio Tis is a special number Aside: Since te Renaissance, many artists and arcitects ave proportioned teir work (e.g., lengt:eigt) to approximate te golden ratio: If (a+b)/a = a/b, ten a = b We will need one special aritmetic fact about : 2 = ((+5 /2 )/2) 2 = ( + 2*5 /2 + 5)/4 = (6 + 2*5 /2 )/4 = ( + 5 /2 )/2 = + ( + 5 /2 )/2 = + /25/29 2

18 Te proof S(-)=, S()=, S()=2 For, S() = +S(-)+S(-2) Teorem: For all, S() > Proof: By induction on Base cases: S() = > = S() = 2 >.62 Inductive case (k > ): Sow S(k+) > k+ assuming S(k) > k and S(k-) > k- S(k+) = + S(k) + S(k-) by definition of S > + k + k- by induction = k + k- by aritmetic (-=) = k- ( + ) by aritmetic (factor k- ) = k- 2 by special property of = k+ by aritmetic (add exponents) /25/29 2

19 Good news Proof means tat if we ave an AVL tree, ten find is O(log n) But as we insert and delete elements, we need to:. Track balance 2. Detect imbalance. Restore balance Is tis tree AVL balanced? How about after insert()? 7 /25/29 22

20 An AVL Tree key value eigt 2 2 cildren /25/29 2

21 AVL tree operations AVL find: Same as BST find AVL insert: First BST insert, ten ceck balance and potentially fix te AVL tree Four different imbalance cases AVL delete: Te easy way is lazy deletion Oterwise, like insert we do te deletion and ten ave several imbalance cases /25/29 24

22 AVL tree insert Let b be te node were an imbalance occurs. Four cases to consider. Te insertion is in te. left subtree of te left cild of b. 2. rigt subtree of te left cild of b.. left subtree of te rigt cild of b. 4. rigt subtree of te rigt cild of b. Idea: Cases & 4 are solved by a single rotation. Cases 2 & are solved by a double rotation. b a c X U V Z /25/

23 Insert: detect potential imbalance. Insert te new node as in a BST (a new leaf) 2. For eac node on te pat from te root to te new leaf, te insertion may (or may not) ave canged te node s eigt. So after recursive insertion in a subtree, detect eigt imbalance and perform a rotation to restore balance at tat node All te action is in defining te correct rotations to restore balance Fact tat makes it a bit easier: Tere must be a deepest element tat is imbalanced after te insert (all descendants still balanced) After rebalancing tis deepest node, every node is balanced So at most one node needs to be rebalanced /25/29 26

24 Case # Example Insert(6) Insert() Insert() /25/29 27

25 Case #: Example Insert(6) Insert() Insert() Tird insertion violates balance property appens to be at te root Wat is te only way to fix tis? /25/29 28

26 Fix: Apply Single Rotation Single rotation: Te basic operation we ll use to rebalance Move cild of unbalanced node into parent position Parent becomes te oter cild (always okay in a BST!) Oter subtrees move in only way BST allows (next slide) AVL Property violated ere 6 2 Single Rotation:. Rotate between self and cild 6 /25/29 29

27 RotateRigt brings up te rigt cild Single Rotation Pseudo-Code root void RotateWitRigt(Node root) { Node temp = root.rigt root.rigt = temp.left temp.left = root root.eigt = max(root.rigt.eigt(), root.left.eigt()) + temp.eigt = max(temp.rigt.eigt(), temp.left.eigt()) + root = temp } X a Y b temp Z /25/29

28 Te example generalized Notational note: Oval: a node in te tree Triangle: a subtree Node imbalanced due to insertion somewere in left-left grandcild increasing eigt of 4 possible imbalance causes (oter tree coming) First we did te insertion, wic would make b imbalanced X + a Y b +2 b +2 Z a + X Y + Z /25/29

29 Te general left-left case Node imbalanced due to insertion somewere in left-left grandcild increasing eigt of 4 possible imbalance causes (oter tree coming) So we rotate at b, using BST facts: X < a < Y < b < Z +2 a + X Y b + a Z + X Y +2 b + Z A single rotation restores balance at te node To same eigt as before insertion (so ancestors now balanced) /25/29 2

30 Anoter example: insert(6) /25/29

31 Anoter example: insert(6) /25/29 4

32 Te general rigt-rigt case Mirror image to left-left case, so you rotate te oter way Exact same concept, but need different code X a + +2 b + b + Y Z X Y a +2 Z + /25/29 5

33 Case # Example Insert() Insert(6) Insert() /25/29 6

34 Two cases to go Unfortunately, single rotations are not enoug for insertions in te left-rigt subtree or te rigt-left subtree Simple example: insert(), insert(6), insert() First wrong idea: single rotation like we did for left-left /25/29 7

35 Two cases to go Unfortunately, single rotations are not enoug for insertions in te left-rigt subtree or te rigt-left subtree Simple example: insert(), insert(6), insert() Second wrong idea: single rotation on te cild of te unbalanced node /25/29 8

36 Sometimes two wrongs make a rigt First idea violated te BST property Second idea didn t fix balance But if we do bot single rotations, starting wit te second, it works! (And not just for tis example.) Double rotation:. Rotate problematic cild and grandcild 2. Ten rotate between self and new cild /25/

37 Double Rotation Pseudo-Code void DoubleRotateWitRigt(Node root) { RotateWitLeft(root.rigt) } RotateWitRigt(root) a First Rotation a c b Z b Z Y c Y X W X /25/29 W 4

38 Double Rotation Completed First Rotation a b a Second Rotation b c Z Y c Y X X Z W W /25/29 4

39 Te general rigt-left case + X X + a a + b V U U b - V c Z b + +2 c Z U X /25/29 42 a - +2 V c + Z

40 Comments X Like in te left-left and rigt-rigt cases, te eigt of te subtree after rebalancing is te same as before te insert So no ancestor in te tree will need rebalancing Does not ave to be implemented as two rotations; can just do: + +2 a +2 b + c + b a c - - U V Z X U V + Z Easier to remember tan you may tink: Move b to grandparent s position and ten put a, c, X, U, V, and Z in te only legal positions for a BST /25/29 4

41 Te last case: left-rigt Mirror image of rigt-left Again, no new concepts, only new code to write X +2 a U + b c + V - Z X + a U b - +2 V c + Z /25/29 44

42 Insert /25/29 45

43 Double rotation, step /25/29 46

44 Double rotation, step /25/29 47

45 Insert, summarized Insert as in a BST Ceck back up pat for imbalance, wic will be of 4 cases: node s left-left grandcild is too tall node s left-rigt grandcild is too tall node s rigt-left grandcild is too tall node s rigt-rigt grandcild is too tall Only one case occurs because tree was balanced before insert After te appropriate single or double rotation, te smallestunbalanced subtree as te same eigt as before te insertion So all ancestors are now balanced /25/29 48

46 Now efficiency Worst-case complexity of find: Tree is balanced Worst-case complexity of insert: Tree starts balanced A rotation is O() and tere s an O(log n) pat to root (Same complexity even witout one-rotation-is-enoug fact) Tree ends balanced Worst-case complexity of buildtree: delete? (see ed. Weiss) requires more rotations: Lazy deletion? /25/29 49

47 Pros and Cons of AVL Trees Arguments for AVL trees:. All operations logaritmic worst-case because trees are always balanced 2. Heigt balancing adds no more tan a constant factor to te speed of insert and delete Arguments against AVL trees:. Difficult to program & debug 2. More space for eigt field. Asymptotically faster but rebalancing takes a little time 4. Most large searces are done in database-like systems on disk and use oter structures (e.g., B-trees, our next data structure) /25/29 5

48 More Examples /25/29 52

49 Insert into an AVL tree: a b e c d /25/29 5 Student Activity

50 Single and Double Rotations: Inserting wat integer values would cause te tree to need a:. single rotation? double rotation? 2 7. no rotation? Student Activity /25/29 54

51 Easy Insert Insert() Unbalanced? /25/29 55

52 Hard Insert Insert() Unbalanced? How to fix? /25/29 56

53 Single Rotation /25/29 57

54 Hard Insert Insert(8) Unbalanced? 7 How to fix? /25/29 58

55 Single Rotation (oops!) /25/29 59

56 Double Rotation (Step #) /25/29 6

57 Double Rotation (Step #2) /25/29 6