CS-102 AVLSv2.pdf
- 1. AVL Trees • binary tree • for every node x, define its balance factor balance factor of x = height of left subtree of x – height of right subtree of x • balance factor of every node x is – 1, 0, or 1 • log2 (n+1) <= height <= 1.44 log2 (n+2)
- 2. Example AVL Search Tree 0 0 0 0 1 0 -1 0 1 0 -1 1 -1 10 7 8 3 1 5 30 40 20 25 35 45 60
- 3. put(9) 0 0 0 0 1 0 -1 0 1 0 -1 1 -1 9 0 -1 0 10 7 8 3 1 5 30 40 20 25 35 45 60
- 4. put(29) 0 0 0 0 1 0 0 1 0 -1 1 -1 10 7 8 3 1 5 30 40 20 25 35 45 60 29 0 -1 -1 -2 RR imbalance => new node is in right subtree of right subtree of white node (node with bf = –2)
- 5. 0 0 0 0 1 0 0 1 0 -1 1 -1 10 7 8 3 1 5 30 40 25 35 45 60 0 RR rotation. 20 0 29 put(29)
- 6. Insert/Put • Following insert/put, retrace path towards root and adjust balance factors as needed. • Stop when you reach a node whose balance factor becomes 0, 2, or –2, or when you reach the root. • The new tree is not an AVL tree only if you reach a node whose balance factor is either 2 or –2. • In this case, we say the tree has become unbalanced.
- 7. A-Node • Let A be the nearest ancestor of the newly inserted node whose balance factor becomes +2 or –2 following the insert. • Balance factor of nodes between new node and A is 0 before insertion.
- 8. Imbalance Types • RR … newly inserted node is in the right subtree of the right subtree of A. • LL … left subtree of left subtree of A. • RL… left subtree of right subtree of A. • LR… right subtree of left subtree of A.
- 9. LL rotation
- 10. LL Rotation • Subtree height is unchanged. • No further adjustments to be done. Before insertion. 1 0 A B BL BR AR h h h A B B’L BR AR After insertion. h+1 h h B A After rotation. BR h AR h B’L h+1 0 0 1 2
- 11. Example
- 12. RR rotation
- 13. In an AVL search After Inserting Node D
- 14. RR rotation
- 15. Example
- 16. LR Rotation (case 1) • Subtree height is unchanged. • No further adjustments to be done. Before insertion. 1 0 A B A B After insertion. C C A After rotation. B 0 -1 2 0 0 0
- 17. LR Rotation (case 2) • Subtree height is unchanged. • No further adjustments to be done. C A CR h-1 AR h 1 0 A B BL CR AR h h-1 h 0 CL h-1 C A B BL CR AR h h-1 h C’L h C B BL h C’L h 1 -1 2 0 -1 0
- 18. LR Rotation (case 3) • Subtree height is unchanged. • No further adjustments to be done. 1 0 A B BL CR AR h h-1 h 0 CL h-1 C A B BL C’R AR h h h CL h-1 C -1 -1 2 C A C’R h AR h B BL h CL h-1 1 0 0
- 19. Example
- 20. RL Rotation
- 23. Single & Double Rotations • Single LL and RR • Double LR and RL LR is RR followed by LL RL is LL followed by RR
- 24. LR Is RR + LL A B BL C’R AR h h h CL h-1 C -1 -1 2 After insertion. A C CL C’R AR h h BL h B 2 After RR rotation. h-1 C A C’R h AR h B BL h CL h-1 1 0 0 After LL rotation.
- 25. Remove An Element 0 0 0 0 1 0 -1 0 1 0 -1 1 -1 10 7 8 3 1 5 30 40 20 25 35 45 60 Remove 8.
- 26. Remove An Element 0 0 0 2 0 -1 0 1 0 -1 1 -1 10 7 3 1 5 30 40 20 25 35 45 60 • Let q be parent of deleted node. • Retrace path from q towards root. q
- 27. New Balance Factor Of q • Deletion from left subtree of q => bf--. • Deletion from right subtree of q => bf++. • New balance factor = 1 or –1 => no change in height of subtree rooted at q. • New balance factor = 0 => height of subtree rooted at q has decreased by 1. • New balance factor = 2 or –2 => tree is unbalanced at q. q
- 28. Imbalance Classification • Let A be the nearest ancestor of the deleted node whose balance factor has become 2 or –2 following a deletion. • Deletion from left subtree of A => type L. • Deletion from right subtree of A => type R. • Type R => new bf(A) = 2. • So, old bf(A) = 1. • So, A has a left child B. bf(B) = 0 => R0. bf(B) = 1 => R1. bf(B) = –1 => R-1.
- 29. R0 Rotation • Subtree height is unchanged. • No further adjustments to be done. • Similar to LL rotation. Before deletion. 1 0 A B BL BR AR h h h B A After rotation. BR h A’R h-1 BL h 1 -1 A B BL BR A’R After deletion. h h h-1 0 2
- 31. R1 Rotation • Subtree height is reduced by 1. • Must continue on path to root. • Similar to LL and R0 rotations. Before deletion. 1 1 A B BL BR AR h h-1 h B A After rotation. BR h-1 A’R h-1 BL h 0 0 A B BL BR A’R After deletion. h h-1 h-1 1 2
- 33. R-1 Rotation • New balance factor of A and B depends on b. • Subtree height is reduced by 1. • Must continue on path to root. • Similar to LR. 1 -1 A B BL CR AR h-1 h b CL C A B BL CR A’R h-1 h-1 CL C b -1 2 C A CR A’R h-1 B BL h-1 CL 0
- 35. Number of Rebalancing Rotations • At most 1 for an insert. • O(log n) for a delete.
- 36. Rotation Frequency • Insert random numbers. No rotation … 53.4% (approx). LL/RR … 23.3% (approx). LR/RL … 23.2% (approx).