Theory & Algorithms 15/01/04. Red-Black Trees. Nicolas Wack, Sylvain Le Groux

Transcription

1 Theory & Algorithms Red-Black Trees 15/01/04 Nicolas Wack, Sylvain Le Groux

2 Balanced search trees Balanced search tree: A search-tree data structure for which a height of O(lg n) is guaranteed when implementing a dynamic set of n items. Examples: AVL trees 2-3 trees trees B-trees Red-black trees

3 Red-black trees This data structure requires an extra onebit color field in each node. Red-black properties: 1. Every node is either red or black. 2. The root and leaves ( s) are black. 3. If a node is red, then its parent is black. 4. All simple paths from any node x to a descendant leaf have the same number of black nodes = black-height(x).

4 Example of a red-black tree h =

5 Example of a red-black tree Every node is either red or black.

6 Example of a red-black tree The root and leaves ( s) are black.

7 Example of a red-black tree If a node is red, then its parent is black.

8 Example of a red-black tree 77 bh = bh = 2 bh = bh = bh = 0 4. All simple paths from any node x to a descendant leaf have the same number of black nodes = black-height(x).

9 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes into their black parents.

10 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes into their black parents.

11 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes into their black parents.

12 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes into their black parents.

13 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes into their black parents.

14 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 lg(n + 1). Proof. INTUITION: Merge red nodes h into their black parents. This process produces a tree in which each node has 2, 3, or 4 children. The tree has uniform depth h of leaves.

15 Proof (continued) We have h h/2, since at most half the leaves on any path are red. The number of leaves in each tree is n + 1 n h' lg(n + 1) h' h/2 h 2 lg(n + 1). h h

16 Query operations Corollary. The queries SEARCH, MIN, MAX, SUCCESSOR, and PREDECESSOR all run in O(lg n) time on a red-black tree with n nodes.

17 Modifying operations The operations INSERT and DELETE cause modifications to the red-black tree: the operation itself, color changes, restructuring the links of the tree via rotations.

18 Rotations BB RIGHT-ROTATE(B) AA αα AA ββ γγ LEFT-ROTATE(A) αα ββ BB γγ Rotations maintain the inorder ordering of keys: a α, b β, c γ a A b B c. A rotation can be performed in O(1) time.

19 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example:

20 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: 33 Insert x =15. Recolor, moving the violation up the tree

21 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: 33 Insert x =15. Recolor, moving the violation up the tree. RIGHT-ROTATE(18)

22 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: 33 Insert x =15. Recolor, moving the violation up the tree. RIGHT-ROTATE(18). LEFT-ROTATE(7) and recolor

23 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: Insert x = Recolor, moving the 33 violation up the tree. RIGHT-ROTATE(18). LEFT-ROTATE(7) and recolor

24 Pseudocode RB-INSERT(T, x) TREE-INSERT(T, x) color[x] RED only RB property 3 can be violated while x root[t] and color[p[x]] = RED do if p[x] = left[p[p[x]] then y right[p[p[x]] y = aunt/uncle of x if color[y] = RED then Case 1 else if x = right[p[x]] then Case 2 Case 2 falls into Case 3 Case 3 else then clause with left and right swapped color[root[t]] BLACK

25 Graphical notation Let All denote a subtree with a black root. s have the same black-height.

26 Case 1 AA x BB CC DD y Recolor AA BB CC new x DD (Or, children of A are swapped.) Push C s black onto A and D, and recurse, since C s parent may be red.

27 Case 2 AA x BB CC y LEFT-ROTATE(A) x AA BB CC y Transform to Case 3.

28 Case 3 x AA BB CC RIGHT-ROTATE(C) y AA BB CC Done! No more violations of RB property 3 are possible.

29 Analysis Go up the tree performing Case 1, which only recolors nodes. If Case 2 or Case 3 occurs, perform 1 or 2 rotations, and terminate. Running time: O(lg n) with O(1) rotations. RB-DELETE same asymptotic running time and number of rotations as RB-INSERT.

30 Deletion into a Red-Black Tree Idea: Same as binary tree but with a fix-up for dealing with colors. (we want to keep the red-black properties) Two Steps: Tree-Delete (T,z) RB-Delete-Fixup(T,x)

31 Reminder: TREE-DELETE(T,z) if left[z] = or right[z] = then y z // z has 0 or 1 child else y TREE-SUCCESSOR(z) if left[y] then x left[y] else x right[y] if x then p[x] p[y] // z has 2 child // y = node to splice // x = child of y or // change of parent for y child

32 TREE-DELETE(T,z) if p[y] = then root[t] x else if y = left[p[y[] then left[p[y]] x else right[p[y]] x // node y has been spliced if y z // spliced node deleted node then key[z] key[y] copy y s satellite data into z return y // calling procedure: free list (h)

33 TREE-DELETE(T,z) - Example if left[z] = or right[z] = then y z // z has 0 or 1 child else y TREE-SUCCESSOR(z) // z has 2 child 15 z

34 TREE-DELETE(T,z) - Example if left[z] = or right[z] = then y z // z has 0 or 1 child else y TREE-SUCCESSOR(z) // z has 2 child 15 z y

35 TREE-DELETE(T,z) - Example if left[y] then x left[y] else x right[y] 15 // x = child of y or z y x

36 TREE-DELETE(T,z) - Example if x then p[x] p[y] // change of parent for y child 15 z y x

37 TREE-DELETE(T,z) - Example if p[y] = then root[t] x else if y = left[p[y[] then left[p[y]] x 15 z y x

38 TREE-DELETE(T,z) - Example if y z return y // spliced node deleted node then key[z] key[y] copy y s satellite data into z 15 // calling procedure: free list z y x 19

39 TREE-DELETE(T,z) - Example if y z return y // spliced node deleted node then key[z] key[y] copy y s satellite data into z 15 // calling procedure: free list z y x 19

40 TREE-DELETE(T,z) - Example if y z return y // spliced node deleted node then key[z] key[y] copy y s satellite data into z 15 // calling procedure: free list z (Free) y x 19

41 Deletion into a Red-Black Tree Fix-Up(T,x): There are 4 different cases. Each case is transformed until no properties are violated anymore FINISH Relations between cases

42 Deletion into a Red-Black Tree Pseudo-code:

43 Deletion into a Red-Black Tree x A B C D w E Case 1 x A B New w D C E Case 2,3,4

44 Deletion into a Red-Black Tree x A B C c D w E Case 2 New x A B C c D E Case 2,4

45 Deletion into a Red-Black Tree x A B C c D w E Case 3 x A B c C New w D E Case 4

46 Deletion into a Red-Black Tree x A B C c D c w E Case 4 A B D C c c E Finished!