1. red-black tree Lai Ah Fur

2. Background: • AVL trees may require many restructure operations (rotations) to be performed after an element removal, • (2,4) trees may require many fusing or split operations to be performed after either an insertion or removal. • The red-black tree, does not have these drawbacks. It requires that only O(1) structural changes be made after an update in order to stay balanced.

3. Property of A red-black tree • A red-black tree is a binary search tree with nodes colored red and black in a way that satisfies the following properties: • Root Property: The root is black. • External Property: Every external node is black. • Internal Property: The children of a red node are black. • Depth property: All the external nodes have the same black depth, which is defined as the number of black ancestors minus one. • The height of a red-black tree storing n items is O(log n).

4. Example of A red-black tree 12 5 15 3 10 13 17 14 7 4 11 6 8

5. x x y α y y x β β β α α γ LEFT-ROTATE(T,x) γ RIGHT-ROTATE(T,y) LEFT-ROTATE(T,x) 1 y←right[x] //Set y. 2 right[x] ← left[y] //Turn y’s left subtree into x’s right subtree. 3 if left[x] ≠ nil[T] 4 thenp[left[y] ] ←x // β’s father 5 p[y ] ← p[x ] //Link x’s parent to y. 6 if p[x ] = nil[T] 7 then root [T] ←y 8 else if x=left[p[x]] 9 thenleft[p[x]] ←y 10 elseright[p[x]] ←y 11 left[y] ←x //Put x on y’s left 12 p[x ] ←y

6. RB-INSERT(T,z) //insert z into T 1 y ← nil[T] 2 x ← root[T] 3 while x≠ nil[T] 4 doy←x 5 ifkey[z]< key[x] 6 then x ← left[x] 7 elsex ← right [x] 8 p[z] ←y 9 if y=nil[T] 10 thenroot[T] ← z 11 elseif key[z]<key[y] 12 thenleft[y] ← z 13 else right[y] ← z 14 left[z] ←nil[T] 15 right[z] ← nil[T] 16 color[z] ← RED 17 RB-INSERT-FIXUT(T,z)

7. RB-INSERT-FIXUP(T,z) 1 whilecolor [p[z]] =RED 2 do ifp[z]=left [p[p[z]]] 3 theny ← right[p[p[z]]] 4 ifcolor [y] =RED 5 thencolor p[z] ←BLACK //case 1 6 color [y ] ←BLACK //case 1 7 color [p[p[z]]] ← RED //case 1 8 z ←[p[z]] //case 1 9 else if z= right [p[z]] 10 then z← p[z] //case 2 11 LEFT-ROTATE(T,z) //case 2 12 color [p[z]] ←BLACK //case 3 13 color [p[p[z]]] ← RED // //case 3 14 RIGHT-ROTATE(T, p[p[z]]) //case 3 15 else (same as then clause with “ right”and “left” exchanged 16 color [root [T]] ←BLACK

8. 11 (a) 2 14 1 7 15 5 8 y 4 z Case 1 : z’s uncle y is red (b) 11 y 2 14 1 7 z 15 5 8 4 Case 2

9. 11 (c) 7 y z 14 2 8 15 1 5 4 Case 3 7 (d) z 2 11 1 5 8 14 15 4 Case 2

10. C C new z A y A D D α z z B B δ ε δ ε β γ β γ C C new z B D y B D y z A γ z A γ ε ε δ δ β α β Case 1-1: z’s uncle y is red α α β γ δε:Black height不變 Problem:連續雙red α 做法:改變parent, uncle, grandparent的color

11. C C y y B B D D α α β β γ γ z z A A δ δ ε ε Case 1-2: z’s uncle y is red new z C y C A D y A D β α z B β z ε ε B δ γ δ γ α β γ δε:Black height不變 If c’s parent is red? Continue… Problem:連續雙red new z 做法:改變parent, uncle, grandparent的color

12. Case 2-1: z’s uncle y is black and z is a right child Case 3-1: z’s uncle y is black and z is a left child Change color: parent & grandpa right rotation Left rotation α C C B A B z A C δ δ y y z A γ B z γ δ α β α β γ β Right rotate C C case 2-1 case 3-1 y B y α α B A z C A δ β B z A z z γ δ α β γ β γ δ Case 2-2 Case 3-2

13. insertion 4 4 4 7 7 (a) 12 (b) (c) Case 3-2 7 7 4 4 12 12 15 (d) (e) Case 1-2 +root must be black

14. 7 7 4 4 12 12 15 3 15 (f) (g)

15. 4 12 3 5 7 15 (h) Insert 5

16. 4 12 3 5 14 7 15 (i)Insert 14 Case 2-2

17. 4 14 3 12 5 7 15 (j)

18. 4 14 3 12 5 18 7 15 (k)

19. 7 4 14 3 5 12 15 18 (l)

20. 16 7 4 14 3 5 12 15 18 (m) Case 2-2

21. 7 4 14 3 5 12 16 15 18 (n)

22. 7 4 14 3 5 12 16 15 18 17 (o) Insert 17 Case 1-2

23. 7 4 14 3 5 12 16 15 18 17 (p) Case 3-2

24. 14 7 16 4 12 15 18 3 5 17 (q)

25. Insertion complexity • The insertion of a key-element item in a red-black tree storing n items can be done in O(log n) time and at most O(log n) recolorings and one trinode restructuring (a restructure operation).

26. RB-DELETE(T,z) 1 ifleft[z]=nil[z] or right[z]=nil[T] 2 then y ←z 3 else z ←TREE-SUCCESSOR(z) 4 ifleft[y] ≠ nil[T] 5 then x← left[y] 6 else x ← right[y] 7 p[x] ← p [y] 8 if p[y]= nil[T] 9 then root [T] ← x 10 else if y=left [p[z]] 11 then left [p[z]] ← x 12 else right [p[z]] ← x 13 if y ≠z 14 then key [z] ← key [y] 15 copy y’s satellite data into z 16 ifcolor [y] = BLACK 17 then RB-DELETE-FIXUP(T,x) 18 return y

27. RB-DELETE FIXUP(T,x) //y為真正被deleted之node, x是y的right or left child • 1 While x ≠ root[T] and color[x] =BLACK • 2 do if x =left [p[x]] • 3 then w ← right [p[x]] • 4 ifcolor[w] = RED • 5 then color[w] ← BLACK //Case 1 • 6 color [p[x]] ← RED //Case 1 • 7 LEFT-ROTATE(T,p[x]) //Case 1 • 8 w ← right [p[x]] //Case 1 • ifcolor [left[w]] = BLACK and color [right[w]]= BLACK • thencolor[w] ← RED //Case 2 • x ← p[x] //Case 2 • 12 else ifcolor [left[w]] = BLACK • 13 then color [left[w]] ← BLACK //Case 3 • 14 color[w] ← RED//Case 3 • 15 RIGHT-ROTATE(T,w) //Case 3 • w ←right [p[x]] //Case 3 • color[w] ← color [p[x]] //Case 4 • color [p[x]] ← BLACK //Case 4 • color [right[w]] ← BLACK //Case 4 • LEFT-ROTATE(T,p[x]) //Case 4 • x ← root[T] //Case 4 • else(same as then clause with”right”and”left”exchanged) • 23 color[x] ← BLACK //若x is red, 改為black, black height 即能維持

28. Case 1 (a) B D restructure x A D w B E w α β C E x A C ε ζ new w γ δ ε ζ α β γ δ (d) (C) Case 2 (b) c new x c B B recolor x A D w A D α β C E α β C E γ δ ε ζ γ δ ε ζ Reduce 1 black height :Red or black y為真正被deleted之node, x是y的right or left child

29. Case 3 c B (c) c B new w A C x x A D w α β γ D α β C E δ E γ δ ε ζ ε ζ Case 4 (d) c c B D x A D w B E α β C c’ E A C c’ ε ζ γ δ ε ζ α β γ δ new x=root[T]

30. Case 1: x’s sibling w is red Case 2: x’s sibling w is black, and both of w’s children are black Case 3: x’s sibling w is black, w’s left child is red, and w’s right child is black Case 4: x’s sibling w is black, and w’s right child is red

31. If v is a 2-node, then keep the (black) children of v as is. • If v is a 3-node, then create a new red node w, give v’s first two (black) children to w, and make w and v;s third child be the two children of v. • If v is a 4-node, then create two new red nodes w and z, give v’s first two (black) children to w, give v’s last two (black) children to z, and make w and z be the two children of v.

32. deletion 14 7 16 4 12 15 18 3 5 17 (a) initial

33. 14 7 16 4 12 15 18 5 17 (b) Delete 3

34. 14 7 16 4 15 18 restructure 5 17 (c) Delete 12

35. 14 5 16 Delete 17 4 7 15 18 17 (d)

36. 14 5 16 Delete 18 4 7 15 18 (e)

37. 14 5 16 4 7 15 (f)

38. 14 5 16 Delete 15 4 7 15 (g) After recoloring

39. 14 5 16 Delete 16 4 7 (h)

40. 5 14 4 4 7 (k) 14 5 14 5 7 4 7 adjustment (j) (i) recoloring

41. 15 15 (a) 13 14 13 14 or 14 13 (b) 7 6 7 8 6 8 (c)

42. Insertion: Case 1: The Sibling w of v is Black. u v w z u u 30 30 v v 20 10 w w z z black 20 10 Double red u 10 10 v 20 30 w z 20 30 (a)

43. 10 30 b 20 a c (b) After a trinode restructuring • Take node z, its parent v, and grandparent u, and temporarily relabel them as a,b,and c, in left-to-right order, so that a, b, c will be visited in this order by an inorder tree traversal. • Replace the grandparent u with the node labeled b, and nodes a and c the children of b, keeping inorder relationships unchanged.

44. 20 40 Case 2: The Sibling w of v is Red. u 30 v 10 20 30 40 w z 10 recoloring (a) u … 30 … 30 … … v w 20 40 10 20 40 z 10 (b)

45. Deletion: case 1: the sibling y of r is black and Has a red child z. x … 30 … 30 … … r y 40 20 10 20 z 10 40 (a)

46. … 30 … x … … 30 y 10 r 10 20 z 40 20 40 (b)

47. 40 b … 20 … 20 … … a c 10 30 10 30 r 40 (a)(b) (c)After restructure

48. Case 2: the sibling y of r is black and both children of y are black. 10 10 30 … x … 30 y 20 r 20 40 40 (a)

49. 10 10 … x … 30 y 20 30 r 20 40 40 After recoloring (b)

50. 30 x 30 y 20 r 20 40 40 (a)