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AVL and Red-Black Trees: Balancing and Insertion/Deletion

Learn about balancing techniques for AVL and Red-Black Trees, as well as insertion and deletion operations.

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AVL and Red-Black Trees: Balancing and Insertion/Deletion

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  1. CS3424 AVL Trees Red-Black Trees

  2. Trees • As stated before, trees are great ways of holding hierarchical data • Insert, Search, Delete ~ O(lgN) • But only if they’re balanced! • So let’s discuss how to assure balance

  3. Rotations x y Left-Rotate(x)   y x Right-Rotate(y)    

  4. AVL Trees (1962) • Uses height property to maintain balance • Height of left & right differ by at most 1 • Uses rotations to restore balance after insertions and deletions

  5. Balancing in AVL Trees • Balance is the difference between the heights of the left and right subtrees • This should range between -1 and 1 • Maintain this balance upon insert and delete • Balance = HR - HL

  6. Left-Left Imbalance 0 Right-Rotate(X) -2 X Y -1 0 Y X      

  7. Left-Right Imbalance -2 -2 X X = +1 +1 Y Y -1      

  8. Left-Right Imbalance (cont) -2 -2 X X = +1 +1 Y Y +1      

  9. Left-Right Imbalance ( left) -2 X +1 Y -1   

  10. Left-Right Imbalance ( left) Left-Rotate(Y) -2 -2 X X +1 -2 Y  -1 0   Y   

  11. Left-Right Imbalance ( left) Right-Rotate(X) -2 0 X  -2 0 +1  X Y 0  Y   

  12. Left-Right Imbalance ( right) -2 X +1 Y +1   

  13. Left-Right Imbalance ( right) Left-Rotate(Y) -2 -2 X X +1 -1 Y  +1 -1   Y   

  14. Left-Right Imbalance ( right) Right-Rotate(X) -2 0  X 0 -1 -1 X  Y -1 Y    

  15. Red-Black Trees (1972/1978) • BST + notion of color (RED / BLACK) • Every node is either red or black • Root is black • Every leaf (NIL) is black • If node red, both children are black • For each node, path to NIL children contains same number of black nodes (black-height)

  16. Possibilities in Growing a Red-Black Tree

  17. Painting a Tree to beRed-Black

  18. Painting a Tree to beRed-Black Root is black

  19. Painting a Tree to beRed-Black Must be black bh=1 bh=1

  20. Painting a Tree to beRed-Black Invalid!

  21. Painting a Tree to beRed-Black Another try!

  22. Painting a Tree to beRed-Black Must be black

  23. Painting a Tree to beRed-Black Must be red

  24. Painting a Tree to beRed-Black bh = 2 or 3 You can finish this with ease!

  25. Inserting & Deleting in Red-Black Trees • Insert similar to binary search trees • “Search” left & right until finding NIL • But now, we need to worry about coloring • Requires possible rotations & “cleanup” • Deleting similar to BSTs as well • Search left & right until finding item • If we removed a black node, “cleanup”

  26. Red-Black Insert • Insert node (z) • z.Color = red • This can cause two possible problems: • Root must be black (initial insert violates this) • Two red nodes cannot be adjacent (this is violated if parent of z is red) • Cleanup

  27. Red-Black Cleanup • y = z’s “uncle” • Three cases: • y is red • y is black and z is a right child • y is black and z is a left child Not mutually exclusive

  28. Case 1 – z’s Uncle is red • y.Color = black • z.Parent.Color = black • z.Parent.Parent.Color = red • z = z.Parent.Parent • Repeat cleanup

  29. Case 1 Visualized 11 2 14 1 7 15 5 8 y y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent repeat cleanup 4 z New Node

  30. Case 1 Visualized 11 2 14 1 7 15 5 8 y y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent repeat cleanup 4 z New Node

  31. Case 1 Visualized 11 2 14 1 7 15 5 8 y y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent repeat cleanup 4 z New Node

  32. Case 1 Visualized 11 2 14 1 7 15 5 8 y y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent repeat cleanup 4 z New Node

  33. Case 1 Visualized 11 2 14 y 1 7 15 z 5 8 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent repeat cleanup 4 New Node

  34. Case 2 – z’s Uncle is black & z is a right child • z = z.Parent • Left-Rotate(T, z) • Do Case 3 • Note that Case 2 is a subset of Case 3

  35. Case 2 Visualized 11 2 14 y 1 7 15 z 5 8 z = z.Parent Left-Rotate(T,z) Do Case 3 4

  36. Case 2 Visualized 11 2 14 y z 1 7 15 5 8 z = z.Parent Left-Rotate(T,z) Do Case 3 4

  37. Case 2 Visualized 11 7 14 y 2 8 15 z 1 5 z = z.Parent Left-Rotate(T,z) Do Case 3 4

  38. Case 3 – z’s Uncle is black & z is a left child • z.Parent.Color = black • z.Parent.Parent.Color = red • Right-Rotate(T, z.Parent.Parent)

  39. Case 3 Visualized 11 7 14 y 2 8 15 z 1 5 z.Parent.Color = black z.Parent.Parent.Color = red Right-Rotate(T, z.Parent.Parent) 4

  40. Case 3 Visualized 11 7 14 y 2 8 15 z 1 5 z.Parent.Color = black z.Parent.Parent.Color = red Right-Rotate(T, z.Parent.Parent) 4

  41. Case 3 Visualized 11 7 14 y 2 8 15 z 1 5 z.Parent.Color = black z.Parent.Parent.Color = red Right-Rotate(T, z.Parent.Parent) 4

  42. Case 3 Visualized 7 2 11 z 8 14 1 5 y 15 4 z.Parent.Color = black z.Parent.Parent.Color = red Right-Rotate(T, z.Parent.Parent)

  43. Analysis of Red-Black Cleanup C New z C A D A D y   B   B   z     Case 1 , , , , and  are all black-rooted and all have same black-height (symmetric L/R on A & B)

  44. Analysis of Red-Black Cleanup C C  y  y A B   B A z z  Case 3    Case 2 , , , and  are all black-rooted and all have same black-height ( is black or this would be case 1) Case 2

  45. Analysis of Red-Black Cleanup C B  y B A C z   A   z  y   Case 3 , , , and  are all black-rooted and all have same black-height

  46. Questions aboutRed-Black Insert • What is the running time? • Why are there only at most 2 rotations in the cleanup? • How many times can the cleanuprepeat (i.e. how many times can we have case 1 in a row)?

  47. Red-Black Delete • Delete as before • But if you take away a black node: • Black height is not “balanced” • Can generate adjacent red nodes • Must perform cleanup • 4 cases • “Beyond the scope of this course…”

  48. FIN

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