Data Structures in Java

Transcription

1 Data Structures in Java Lecture 10: AVL Trees. 10/1/015 Daniel Bauer

2 Balanced BSTs Balance condition: Guarantee that the BST is always close to a complete binary tree (every node has exactly two or zero children). Then the height of the tree will be O(log N) and all BST operations will run in O(log N).

3 AVL Tree Condition An AVL Tree is a Binary Search Tree in which the following balance condition holds after each operation: For every node, the height of the left and right subtree differs by at most not an AVL tree

4 AVL Trees Height of an AVL tree is at most ~ 1.44 log(n+)-1.38 = O(log N) How to maintain the balance condition? Rebalance the tree after each change (insertion or deletion). Rebalancing must be cheap.

5 Outside Imbalance node k violates the balance condition k k k1 z x k1 x y y z left subtree of left child too high right subtree of right child too high Solution: Single rotation

6 Inside Imbalance node k violates the balance condition k k k1 z x k1 x y Y z right subtree of left child too high left subtree of right child too high Solution: Double rotation

7 Single Rotation k x k1 y z Maintain BST property: x is still left subtree of k1. z is still right subtree of k. For all values v in y: k1 < v < k so y becomes new left subtree of k.

8 Single Rotation k1 k x y z Maintain BST property: x is still left subtree of k1. z is still right subtree of k. For all values v in y: k1 < v < k so y becomes new left subtree of k.

9 Single Rotation x k1 y k z Maintain BST property: x is still left subtree of k1. Modify 3 references: k.left = k1.right k1.right = k parent(k).left = k1 or parent(k).right = k1 or z is still right subtree of k. For all values v in y: k1 < v < k so y becomes new left subtree of k.

10 Maintaining Balance in an AVL Tree Assume the tree is balanced. After each insertion, find the lowest node k that violates the balance condition (if any). Perform rotation to re-balance the tree. Rotation maintains original height of subtree under k before the insertion. No further rotations are needed.

11 Single Rotation Example insert(3) 3

12 Single Rotation Example insert(3) insert() 3

13 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 3

14 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 3

15 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 3 insert(4) 4

16 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 3 insert(4) 4 insert(5) rotate_right(3) 5

17 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 4 insert(4) 3 5 insert(5) rotate_right(3)

18 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 1 4 insert(4) 3 5 insert(5) insert(6) rotate_right(3) rotate_right() 6

19 Single Rotation Example insert(3) 4 insert() insert(1) rotate_left(3) 5 insert(4) insert(5) rotate_right(3) insert(6) rotate_right()

20 Single Rotation Example insert(3) 4 insert() insert(1) rotate_left(3) 5 insert(4) insert(5) insert(6) rotate_right(3) rotate_right() 7 insert(7) rotate_right(5)

21 Single Rotation Example insert(3) insert() insert(1) rotate_left(3) 4 insert(4) insert(5) insert(6) rotate_right(3) rotate_right() insert(7) rotate_right(5)

22 Single Rotation does not work for Inside Imbalance k k1 z x y

23 Single Rotation does not work for Inside Imbalance k1 x y k z Result is not an AVL tree. Now k1 is violates the balance condition. Problem: Tree y cannot move and it is too high.

24 Double Rotation (1) y is non-empty (imbalance due to insertion into y or deletion from z) so we can view y as a root and two sub-trees. k3 k1 z x y

25 Double Rotation (1) y is non-empty (imbalance due to insertion into y or deletion from z) so we can view y as a root and two sub-trees. k3 k1 z x k yl yr either yl or yr is two levels deeper than z (or both are empty).

26 Double Rotation () k3 k1 z Maintain BST property: x is still left subtree of k1. z is still right subtree of k3. x yl k yr For all values v in yl: k1 < v < k so yl becomes new right subtree of k1. For all values w in yr: k < w < k3 so yr becomes new left subtree of k3.

27 Double Rotation () k k1 k3 Maintain BST property: x yl yr z x is still left subtree of k1. z is still right subtree of k3. For all values v in yl: k1 < v < k so yl becomes new right subtree of k1. For all values w in yr: k < w < k3 so yr becomes new left subtree of k3.

28 Double Rotation () These are actually two single rotations: First at k1, then at k3. k3 k1 z x k yl yr

29 Double Rotation () These are actually two single rotations: First at k1, then at k3. k3 k z k1 yr x yl

30 Double Rotation () These are actually two single rotations: First at k1, then at k3. k k1 k3 x yl yr z

31 Double Rotation (3) Modify 5 references: k parent(k3).left = k or parent(k3).right = k k1 k3 k.left = k1 k.right = k3 k1.right = root(yl) x yl yr z k3.left = root(yr)

32 Double Rotation Example

33 Double Rotation Example insert(16)

34 Double Rotation Example insert(16) insert(7) rotate(7)

35 Double Rotation Example insert(16) insert(7) rotate(7) 4 6 k x k3 16 k z 15 yl yr

36 Double Rotation Example insert(16) insert(7) rotate(7) 4 insert(14) rotate(6)

37 Double Rotation Example insert(16) insert(7) rotate(7) 4 insert(14) rotate(6) 6 k x k k z 14 yl yr

38 Double Rotation Example insert(16) insert(7) rotate(7) 4 insert(14) rotate(6)