CS 3424 AVL Trees RedBlack Trees Trees As
CS 3424 AVL Trees Red-Black Trees
Trees • As stated before, trees are great ways of holding hierarchical data • Insert, Search, Delete ~ O(lg. N) • But only if they’re balanced! • So let’s discuss how to assure balance
Rotations x y Left-Rotate(x) y x Right-Rotate(y)
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
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
Left-Left Imbalance -2 -1 Right-Rotate(X) X 0 Y Y X 0
Left-Right Imbalance -2 +1 -2 X = Y +1 -1 Y X
Left-Right Imbalance (cont) -2 +1 -2 X = Y +1 Y X
Left-Right Imbalance ( left) -2 +1 Y -1 X
Left-Right Imbalance ( left) -2 +1 Y Left-Rotate(Y) X -2 -2 -1 Y 0 X
Left-Right Imbalance ( left) -2 -2 Y 0 Right-Rotate(X) X 0 X +1 Y 0
Left-Right Imbalance ( right) -2 +1 Y +1 X
Left-Right Imbalance ( right) -2 +1 Y X Left-Rotate(Y) -2 -1 +1 -1 Y X
Left-Right Imbalance ( right) -2 -1 -1 Y Right-Rotate(X) X -1 Y X 0
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)
Possibilities in Growing a Red. Black Tree
Painting a Tree to be Red-Black
Painting a Tree to be Red-Black Root is black
Painting a Tree to be Red-Black Must be black bh=1
Painting a Tree to be Red-Black
Painting a Tree to be Red-Black
Painting a Tree to be Red-Black Must be black
Painting a Tree to be Red-Black Must be red
Painting a Tree to be Red-Black bh = 2 or 3
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”
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
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
Case 1 – z’s Uncle is red • • y. Color = black z. Parent. Color = red z = z. Parent • Repeat cleanup
Case 1 Visualized 11 2 14 1 7 5 15 8 y z 4 New Node y. Color = black z. Parent. Color = red z = z. Parent repeat cleanup
Case 1 Visualized 11 2 14 1 7 5 15 8 y z 4 New Node y. Color = black z. Parent. Color = red z = z. Parent repeat cleanup
Case 1 Visualized 11 2 14 1 7 5 15 8 y z 4 New Node y. Color = black z. Parent. Color = red z = z. Parent repeat cleanup
Case 1 Visualized 11 2 14 1 7 5 15 8 y z 4 New Node y. Color = black z. Parent. Color = red z = z. Parent repeat cleanup
Case 1 Visualized 11 14 y 2 1 z 7 5 15 8 4 New Node y. Color = black z. Parent. Color = red z = z. Parent repeat cleanup
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
Case 2 Visualized 11 14 y 2 1 z 7 5 4 15 8 z = z. Parent Left-Rotate(T, z) Do Case 3
Case 2 Visualized 11 14 y z 2 1 7 5 4 15 8 z = z. Parent Left-Rotate(T, z) Do Case 3
Case 2 Visualized 11 14 y 7 z 2 8 1 5 4 15 z = z. Parent Left-Rotate(T, z) Do Case 3
Case 3 – z’s Uncle is black & z is a left child • z. Parent. Color = black • z. Parent. Color = red • Right-Rotate(T, z. Parent)
Case 3 Visualized 11 14 y 7 z 2 8 1 15 5 4 z. Parent. Color = black z. Parent. Color = red Right-Rotate(T, z. Parent)
Case 3 Visualized 11 14 y 7 z 2 8 1 15 5 4 z. Parent. Color = black z. Parent. Color = red Right-Rotate(T, z. Parent)
Case 3 Visualized 11 14 y 7 z 2 8 1 15 5 4 z. Parent. Color = black z. Parent. Color = red Right-Rotate(T, z. Parent)
Case 3 Visualized 7 z 2 11 1 5 4 8 14 y 15 z. Parent. Color = black z. Parent. Color = red Right-Rotate(T, z. Parent)
Analysis of Red-Black Cleanup New z C A D z B A y C D B Case 1 , , and are all black-rooted and all have same black-height (symmetric L/R on A & B)
Analysis of Red-Black Cleanup C C y A z B Case 2 A z y B Case 3 Case 2 , , , and are all black-rooted and all have same black-height ( is black or this would be case 1)
Analysis of Red-Black Cleanup C y B A z C y Case 3 , , , and are all black-rooted and all have same black-height
Questions about Red-Black Insert • What is the running time? • Why are there only at most 2 rotations in the cleanup? • How many times can the cleanup repeat (i. e. how many times can we have case 1 in a row)?
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…”
FIN
- Slides: 48