CS 3343 Fall 2007 Red-black trees Carola Wenk
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1 CS 3343 Fall 2007 Red-black trees Carola Wenk Slides courtesy of Charles Leiserson with small changes by Carola Wenk CS 334 Analysis of Algorithms 1
2 Search Trees A binary search tree is a binary tree. Each node stores a key. The tree fulfills the binary search tree property: For every node x holds: left(x) x, if x s left child left(x) exists x right(x), if x s right child right(x) exists CS 334 Analysis of Algorithms 2
3 Search Trees Different variants of search trees: Balanced search trees (guarantee height of log n for n elements) k-ary search trees (such as B-trees, 2-3-trees) Search trees that store the keys only in the leaves, and store additional split-values in the internal nodes CS 334 Analysis of Algorithms 3
4 ADT Dictionary / Dynamic Set Abstract data type (ADT) Dictionary (also called Dynamic Set): A data structure which supports operations 7 Insert Delete Find Using balanced binary search trees we can implement a dictionary data structure such that each operation takes O(log n) time. CS 334 Analysis of Algorithms 4
5 Balanced search trees Balanced search tree: A search-tree data structure for which a height of O(log n) is guaranteed when implementing a dynamic set of n items. Examples: AVL trees 2-3 trees trees B-trees Red-black trees CS 334 Analysis of Algorithms 5
6 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 is black. 3. The leaves ( s) are black. 4. If a node is red, then both its children are black. 5. All simple paths from any node x, excluding x, to a descendant leaf have the same number of black nodes = black-height(x). CS 334 Analysis of Algorithms 6
7 Example of a red-black tree h = CS 334 Analysis of Algorithms 7
8 Example of a red-black tree Every node is either red or black. CS 334 Analysis of Algorithms 8
9 Example of a red-black tree , 3. The root and leaves ( s) are black. CS 334 Analysis of Algorithms 9
10 Example of a red-black tree If a node is red, then both its children are black. CS 334 Analysis of Algorithms 10
11 Example of a red-black tree 77 bh = bh = 2 bh = bh = bh = 0 5. All simple paths from any node x, excluding x, to a descendant leaf have the same number of black nodes = black-height(x). CS 334 Analysis of Algorithms 11
12 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) INTUITION: Merge red nodes into their black parents. CS 334 Analysis of Algorithms 12
13 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) INTUITION: Merge red nodes into their black parents. CS 334 Analysis of Algorithms 13
14 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) INTUITION: Merge red nodes into their black parents. CS 334 Analysis of Algorithms 14
15 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) INTUITION: Merge red nodes into their black parents. CS 334 Analysis of Algorithms 15
16 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) INTUITION: Merge red nodes into their black parents. CS 334 Analysis of Algorithms 16
17 Height of a red-black tree Theorem. A red-black tree with n keys has height h 2 log(n + 1). Proof. (The book uses induction. Read carefully.) 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. CS 334 Analysis of Algorithms 17
18 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' log(n + 1) h' h/2 h 2 log(n + 1). h h CS 334 Analysis of Algorithms 18
19 Query operations Corollary. The queries SEARCH, MIN, MAX, SUCCESSOR, and PREDECESSOR all run in O(log n) time on a red-black tree with n nodes CS 334 Analysis of Algorithms 19
20 Modifying operations The operations INSERT and DELETE cause modifications to the red-black tree: 1. the operation itself, 2. color changes, 3. restructuring the links of the tree via rotations. CS 334 Analysis of Algorithms 20
21 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. Rotations maintain the binary search tree property A rotation can be performed in O(1) time. CS 334 Analysis of Algorithms 21
22 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 is black. 3. The leaves ( s) are black. 4. If a node is red, then both its children are black. 5. All simple paths from any node x, excluding x, to a descendant leaf have the same number of black nodes = black-height(x). CS 334 Analysis of Algorithms 22
23 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: Insert x = CS 334 Analysis of Algorithms 23
24 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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 CS 334 Analysis of Algorithms 24
25 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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 CS 334 Analysis of Algorithms 25
26 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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) CS 334 Analysis of Algorithms 26
27 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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) CS 334 Analysis of Algorithms 27
28 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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) CS 334 Analysis of Algorithms 28
29 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring. Example: Insert x =15. Recolor, moving the violation up the tree. RIGHT-ROTATE(18). LEFT-ROTATE(7) CS 334 Analysis of Algorithms 29
30 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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 CS 334 Analysis of Algorithms 30
31 Insertion into a red-black tree IDEA: Insert x in tree. Color x red. Only redblack property 4 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 CS 334 Analysis of Algorithms 31
32 Pseudocode RB-INSERT(T, x) TREE-INSERT(T, x) color[x] RED only RB property 4 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 CS 334 Analysis of Algorithms 32
33 Graphical notation Let All denote a subtree with a black root. s have the same black-height. CS 334 Analysis of Algorithms 33
34 Case 1 AA x BB CC DD y Recolor AA BB CC Recurse new x DD (Or, A s children are swapped.) p[x] = left[p[p[x]] y = right[p[p[x]] color[y] = RED Push C s black onto A and D, and recurse, since C s parent may be red. CS 334 Analysis of Algorithms 34
35 Case 2 AA x BB CC y LEFT-ROTATE(A) x AA BB CC y p[x] = left[p[p[x]] y = right[p[p[x]] color[y] = BLACK x = right[p[x]] Transform to Case 3. CS 334 Analysis of Algorithms 35
36 Case 3 x AA BB CC RIGHT-ROTATE(C) y (and recolor) AA BB CC p[x] = left[p[p[x]] y = right[p[p[x]] color[y] = BLACK x = left[p[x]] Done! No more violations of RB property 4 are possible. CS 334 Analysis of Algorithms 36
37 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(log n) with O(1) rotations. RB-DELETE same asymptotic running time and number of rotations as RB-INSERT (see textbook). CS 334 Analysis of Algorithms 37
38 Pseudocode (part II) else then clause with left and right swapped p[x] = right[p[p[x]] then y left[p[p[x]] y = aunt/uncle of x if color[y] = RED then Case 1 else if x = left[p[x]] then Case 2 Case 2 falls into Case 3 Case 3 color[root[t]] BLACK CS 334 Analysis of Algorithms 38
39 Case 1 DD y CC AA x BB Recolor DD y CC new x AA x BB (Or, A s children are swapped.) p[x] = right[p[p[x]] y = left[p[p[x]] color[y] = RED Push C s black onto A and D, and recurse, since C s parent may be red. CS 334 Analysis of Algorithms 39
40 Case 2 y x CC BB AA RIGHT-ROTATE(A) y CC B B x AA p[x] = right[p[p[x]] y = left[p[p[x]] color[y] = BLACK x = left[p[x]] Transform to Case 3. CS 334 Analysis of Algorithms 40
41 Case 3 CC y B B x LEFT-ROTATE(C) (and recolor) AA BB CC AA p[x] = right[p[p[x]] y = left[p[p[x]] color[y] = BLACK x = right[p[x]] Done! No more violations of RB property 4 are possible. CS 334 Analysis of Algorithms 41
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