1. Min Chen School of Computer Science and Engineering Seoul National University Data Structure: Chapter 8

2.  Motivation of Red-Black Trees  Balanced and Unbalanced Trees  AVL Trees  Definition of Red-Black Trees  Operations for Red-Black Trees  Search  Insertion

3. 2020 3030 3535 2525 1010 1515 5  Balanced and Unbalanced Trees

4.  AVL-Tree  The balance factor of a node is the height of its left subtree minus the height of its right subtree  A node with balance factor 1, 0, or -1 is considered balanced 6060 4040 5050 3030 Balance Factor: 2-0=2 Balance Factor: 2-0=2 Balance Factor: 0-1=-1 Balance Factor: 0-1=-1 4545 Unbalanced

5.  AVL-Tree: Rotation Operation

6. Left Right Case 6060 4040 5050 3030 4545 6060 4040 5050 3030 4545 6060 4040 5050 3030 4545

7.  Definition: a binary tree, satisfying: 1. Every node is colored either red or black 2. The root is black 3. If a node is red, its children must be black ▪ consecutive red nodes are disallowed 4. Every path from a node to a null reference must contain the same number of black nodes

8.  The insertion sequence is  10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40, 5, 55 30 15 550 10 70 8560 4055 809065 20

9. Each path must contain the same number of black nodes. (Rule #4) Consecutive red nodes are not allowed. (Rule #3) The longest path is at most twice the length of the shortest path

10.  B = total black nodes from root to leaf  N = total all nodes  H = height All operations guaranteed logarithmic!

11.  A new node must be colored red  Why? ▪ A new item is always inserted as a leaf in the tree ▪ If we color a new item black, then the number of black nodes from root would be different (violate property #4)  If the parent is black, no problem.  If the parent is red, we create two consecutive red nodes (violate property #3) ▪ Thus, we have to do some rotating/recolouring…

12. AB G X S P C DE AB G X S P C DE X: new node P: parent S: sibling G: Grandparent  Case after insertion:  Consecutive red (P & X)  Sibling of parent (S) is black  X is “outer node” (left-left or right-right)

13. BC G X S P A DE AB G P S X C DE X: new node P: parent S: sibling G: Grandparent Case after insertion: Consecutive red (P & X) Sibling of parent (S) is black X is “inner node” (left-right or left)

14.  Case after insertion:  Consecutive red  Sibling of parent is red  Outer node (left-left or right-right) AB G X SP C DE AB G X S P C DE But what if P’s parent is red?  We have to keep going up the tree all the way to the root 

15.  The solution: prevent S from ever being red!  Starting from the root (searching for insertion point)  Never allow 2 red siblings  If we see a node X with 2 red children, do a colour flip. X C2C1 X C2

16.  Maintains property #4  Possible violation of #3: if X’s parent is red! ▪ Do single or double rotation ▪ X’s parent’s sibling can never be red!  Set the root to black (to maintain property #2) X C2C1 X C2

17. AB G X SP C DE AB G X SP C DE If we do the colour flipping on the way down to the insertion point, we will never reach a condition where P & S are red!

18. 30 15 5 10 70 85 60 50 55 8090 65 20 40 18

19. 30 15 5 10 70 85 60 50 55 8090 65 20 40 18 2

20. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10

21. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10 45

22. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10 45 Color-flip!

23. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10 45 Color-flip!

24. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10 45

25. 30 15 2 5 70 85 60 50 55 8090 65 20 40 18 10 45

26.  The insertion sequence is 10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40, 5, 55

27.  Deletion in BST: only leaf nodes or nodes with one child are really deleted (Why?)  If the deleted node is red: no problem (all properties maintained). Leaf nodes: Single child nodes:

28.  If node to be deleted is black  violate property #4  Always ensure that the node to be deleted is red.  Top-down traversal from root (looking for node to be deleted): X: visited node P: parent S: sibling B P S X A CD Idea: make sure that X is red!

29.  P is red (inductive invariant)  X and S are black (result of property #3)  2 cases:  1. Both X’s children (A & B) are black  2. X has at least one red child (A, B, or both) B P S X A CD

30.  Depends on children of S (C & D):  Both C & D are black: simply colour-flip: B P S X A C D B P S X A C D

31. Case 1: Both X’s children are black Outer child of S (C) is Red: do a single rotation and recolour B P S X A C D C S D P B X A

32.  Inner child of S (C) is Red: do a double rotation and recolour B P S X A D C C S P B X A D

33.  Recurse down to X’s child  If we land on a red node, fine.  If we land on a black node, rotate sibling and parent: B P S X A C C S P B X A D D

34.  Red-Black trees use color as balancing information instead of height in AVL trees.  An insertion may cause a local perturbation (two consecutive red nodes)  The pertubation is either  resolved locally (rotations), or  propagated to a higher level in the tree by recoloring (color flip)  O(1) for a rotation or color flip  At most one restructuring per insertion.  O(log n) color flips  Total time: O(log n)