2. Bottom-Up Red-Black Trees Top-down red-black trees require O(log n) rotations per insert/delete. Color flips cheaper than rotations. Priority search trees.  Two keys per element.  Search tree on one key, priority queue on other.  Color flip doesn’t disturb priority queue property.  Rotation disturbs priority queue property.  O(log n) fix time per rotation => O(log 2 n) overall time.

3. Bottom-Up Red-Black Trees Bottom-up red-black tree properties.  At most 1 rotation per insert/delete.  O(1) amortized complexity to restructure following an insert/delete.

4. Bottom-Up Insert New pair is placed in a new node, which is inserted into the red-black tree. New node color options.  Black node => one root-to-external-node path has an extra black node (black pointer).  Hard to remedy.  Red node => one root-to-external-node path may have two consecutive red nodes (pointers).  May be remedied by color flips and/or a rotation.

5. Classification Of 2 Red Nodes/Pointers ab c d gp pp p XYz  X => relationship between gp and pp.  pp left child of gp => X = L.  Y => relationship between pp and p.  p right child of pp => Y = R.  z = b (black) if d = null or a black node.  z = r (red) if d is a red node.

6. XYr Color flip. ab c d gp pp p ab c d gp pp p Move p, pp, and gp up two levels. Continue rebalancing if necessary.

7. LLb Rotate. Done! Same as LL rotation of AVL tree. y x ab z cd ab c d gp pp p x y z

8. LRb Rotate. Done! Same as LR rotation of AVL tree. RRb and RLb are symmetric. y x ab z cd bc a d gp pp p y x z

9. Exercise: build the red-black tree Build a red-black tree from the following key values in sequence: 45, 28, 36, 22, 68, 50, 10, 98, 26, 15, 25, 27, 78

10. Delete Delete as for unbalanced binary search tree. If red node deleted, no rebalancing needed. If black node deleted, a subtree becomes one black pointer (node) deficient.

11. Delete A Black Leaf 10 7 8 1 5 30 40 20 25 35 45 60 3 Delete 8.

12. Delete A Black Leaf y y is root of deficient subtree. py is parent of y. 10 7 1 5 30 40 20 25 35 45 60 3 py

13. Delete A Black Degree 1 Node 10 7 8 1 5 30 40 20 25 35 45 60 3 Delete 45. y y is root of deficient subtree. py

14. Delete A Black Degree 2 Node 10 7 8 1 5 30 40 20 25 35 45 60 3 Not possible, degree 2 nodes are never deleted.

15. Rebalancing Strategy If y is a red node, make it black. 10 7 8 1 5 30 40 20 25 35 45 60 3 y py

16. Rebalancing Strategy Now, no subtree is deficient. Done! 60 10 7 8 1 5 30 40 20 25 35 45 3 y py

17. Rebalancing Strategy y is a black root (there is no py). Entire tree is deficient. Done! 60 10 7 8 1 5 30 40 20 25 35 45 3 y

18. Rebalancing Strategy y is black but not the root (there is a py). Xcn  y is right child of py => X = R.  Pointer to v is black => c = b.  v has 1 red child => n = 1. ab y py v

19. Rb0 (case 1) Color change. Now, py is root of deficient subtree. Continue! ab y py v y ab v

20. Rb0 (case 2) Color change. Deficiency eliminated. Done! ab y py v y ab v

21. Rb1 (case 1) LL rotation. Deficiency eliminated. Done! ab y py v a by v

22. Rb1 (case 2) LR rotation. Deficiency eliminated. Done! a y py v bc wcy w ab v

23. Rb2 LR rotation. Deficiency eliminated. Done! a y py v bc wcy w ab v

24. Rr(n) n = # of red children of vs right child w. a y py v bc w

25. Rr(0) LL rotation. Done! ab y py v a by v

26. Rr(1) (case 1) LR rotation. Deficiency eliminated. Done! a y py v b w c yc w a v b

27. Rr(1) (case 2) Rotation. Deficiency eliminated. Done! a y py v b w cd x yd x a v bc w

28. Rr(2) Rotation. Deficiency eliminated. Done! a y py v b w cd x dy x a v bc w