1. Red-Black Trees—Again
rank(x) = # black pointers on path from x to an external node.
Same as #black nodes (excluding x) from x to an external node.
rank(external node) = 0.

2. An Example 3 10 7 8 1 5 30 40 20 25 35 45 60 3 2 2 1 1 1 2 1 1 1 1 1 1

3. Properties Of rank(x) rank(x) = 0 for x an external node.10 7 8 1 5 30 40 20 25 35 45 60 3 2
rank(x) = 0 for x an external node.
rank(x) = 1 for x parent of external node.

4. Properties Of rank(x) 10 7 8 1 5 30 40 20 25 35 45 60 3 2
p(x) exists => rank(x) <= rank(p(x)) <= rank(x) + 1.
g(x) exists => rank(x) < rank(g(x)).

5. Red-Black Tree
A binary search tree is a red-black tree iff integer ranks can be assigned to its nodes so as to satisfy the stated 4 properties of rank.

6. Relationship Between rank() And Color10 7 8 1 5 30 40 20 25 35 45 60 3 2
(p(x),x) is a red pointer iff rank(x) = rank(p(x)).
(p(x),x) is a black pointer iff rank(x) = rank(p(x)) – 1.

7. Relationship Between rank() And Color
Root is black.
Other nodes:
Red iff pointer from parent is red.
Black iff pointer from parent is black.

8. Relationship Between rank() And Color10 7 8 1 5 30 40 20 25 35 45 60 3 2
Given rank(root) and node/pointer colors, remaining ranks may be computed on way down.

9. rank(root) & tree height10 7 8 1 5 30 40 20 25 35 45 60 3 2
Height ≤ 2 * rank(root).

10. rank(root) & tree height10 7 8 1 5 30 40 20 25 35 45 60 3 2
No external nodes at levels 1, 2, …, rank(root).

11. rank(root) & tree height
No external nodes at levels 1, 2, …, rank(root).
So, #nodes ≥ S1 ≤ i ≤ rank(root) 2i -1 = 2 rank(root) – 1.
So, rank(root) ≤ log2(n+1).
So, height(root) ≤ 2log2(n+1).

12. Join(S,m,B) Input Output Dictionary S of pairs with small keys.
Dictionary B of pairs with big keys.
An additional pair m.
All keys in S are smaller than m.key.
All keys in B are bigger than m.key.
Output
A dictionary that contains all pairs in S and B plus the pair m.
Dictionaries S and B may be destroyed.

13. Join Binary Search Treesm B
O(1) time.

14. Join Red-black Trees
When rank(S) = rank(B), use binary search tree method. S m B
rank(root) = rank(S) + 1 = rank(B) + 1.

15. rank(S) > rank(B)
Follow right child pointers from root of S to first node x whose rank equals rank(B). a b x
p(x) m B a b x
p(x) S
If p(x) is a red node, we have 2 red nodes in a sequence. Retrace to root fixing 2 consecutive red nodes problem.

16. rank(S) > rank(B)
If there are now 2 consecutive red pointers/nodes, perform bottom-up rebalancing beginning at m.
O(rank(S) – rank(B)). a b x
p(x) m B a b x
p(x) S

17. rank(S) < rank(B)
Follow left child pointers from root of B to first node x whose rank equals rank(S).
Similar to case when rank(S) > rank(B).

18. Split(k) Inverse of join. Obtain
S … dictionary of pairs with key < k.
B … dictionary of pairs with key > k.
m … pair with key = k (if present).

19. Split A Binary Search Treed D e E f m g

20. Split A Binary Search TreeD c E d m e f g

21. Split A Binary Search TreeD c E d m e f g

22. Split A Binary Search TreeD c E d m e f g

23. Split A Binary Search TreeD b C d c E m e f g

24. Split A Binary Search TreeD b C d c E e m f g

25. Split A Binary Search TreeD b C g d c E e f m

26. Split A Red-Black Tree
Previous strategy does not split a red-black tree into two red-black trees.
Must do a search for m followed by a traceback to the root.
During the traceback use the join operation to construct S and B.

27. Split A Red-Black Tree
S = f
B = g b A a B c C d D e E f m g

28. Split A Red-Black Tree
S = f
B = g b A a B c C d D e E

29. Split A Red-Black Tree S = f B = g S = join(e, E, S) b A a B c C d D e

30. Split A Red-Black Tree S = f B = g S = join(e, E, S) B = join(B, D, d)

31. Split A Red-Black Tree S = f B = g S = join(e, E, S) B = join(B, D, d)
S = join(c, C, S) c

32. Split A Red-Black Tree S = f B = g S = join(e, E, S) B = join(B, D, d)
S = join(c, C, S)
B = join(B, B, b)

33. Split A Red-Black Tree S = f B = g S = join(e, E, S) B = join(B, D, d)
S = join(c, C, S)
B = join(B, B, b)
S = join(a, A, S)

34. Complexity Of Split
O(log n)