Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Published byHoratio Eaton Modified over 9 years ago

Similar presentations

Presentation on theme: "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."— Presentation transcript:

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 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

3 Properties Of rank(x) rank(x) = 0 for x an external node. rank(x) = 1 for x parent of external node. 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

4 Properties Of rank(x) p(x) exists => rank(x) <= rank(p(x)) <= rank(x) + 1. g(x) exists => rank(x) < rank(g(x)). 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

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 Color (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. 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

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 Color Given rank(root) and node/pointer colors, remaining ranks may be computed on way down. 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

9 rank(root) & tree height Height <= 2 * rank(root). 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

10 rank(root) & tree height No external nodes at levels 1, 2, …, rank(root). 10 7 8 1 5 30 40 20 25 35 45 60 3 0000 00 0 00 0 0 0 00 1 1 1 1 1 1 1 1 2 1 2 2 3

11 rank(root) & tree height No external nodes at levels 1, 2, …, rank(root).  So, #nodes >=  1 <= i <= rank(root) 2 i -1 = 2 rank(root) – 1.  So, rank(root) <= log 2 (n+1). So, height(root) <= 2log 2 (n+1).

12 Join(S,m,B) Input  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 Trees O(1) time. S m B

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). ab x p(x) S ab x m B

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)). ab x p(x) S ab x m B

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 Tree b A a B c C d D e E f m g SB

20 b A a B c C d D e E f m g BS

21 b A a B c C d D e E f m g BS

22 b A a B c C d D e E f m g BS

23 b A a B c C d D e E f m g BS

24 b A a B c C d D e E f m g BS

25 b A a B c C d D e E f m g BS

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 b A a B c C d D e E f m g S = f B = g

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

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

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

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

32 Split A Red-Black Tree b A a B 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 A a 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)

Download ppt "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."

Similar presentations

Splay Trees Binary search trees.

Splay Trees Binary search trees.
Part II. Delete an Node from an AVL Tree Consider to delete 4 10 2 1 8 1 0 9 5 0 3 1 6 4 0 7 0 11 12 0.

Part II. Delete an Node from an AVL Tree Consider to delete
Binary Tries (continued) split(k). Similar to split algorithm for unbalanced binary search trees. Construct S and B on way down the trie. Follow with a.

Binary Tries (continued) split(k). Similar to split algorithm for unbalanced binary search trees. Construct S and B on way down the trie. Follow with a.
AA Trees another alternative to AVL trees. Balanced Binary Search Trees A Binary Search Tree (BST) of N nodes is balanced if height is in O(log N) A balanced.

AA Trees another alternative to AVL trees. Balanced Binary Search Trees A Binary Search Tree (BST) of N nodes is balanced if height is in O(log N) A balanced.
A balanced life is a prefect life.

A balanced life is a prefect life.
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.

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.
Chapter 6: Transform and Conquer 2-3-4 Trees, Red-Black Trees The Design and Analysis of Algorithms.

Chapter 6: Transform and Conquer Trees, Red-Black Trees The Design and Analysis of Algorithms.
Digital Search Trees & Binary Tries Analog of radix sort to searching. Keys are binary bit strings.  Fixed length – 0110, 0010, 1010, 1011.  Variable.

Digital Search Trees & Binary Tries Analog of radix sort to searching. Keys are binary bit strings.  Fixed length – 0110, 0010, 1010,  Variable.
1 /26 Red-black tree properties Every node in a red-black tree is either black or red Every null leaf is black No path from a leaf to a root can have two.

1 /26 Red-black tree properties Every node in a red-black tree is either black or red Every null leaf is black No path from a leaf to a root can have two.
1 /26 Red-black tree properties Every node in a red-black tree is either black or red Every null leaf is black No path from a leaf to a root can have two.

1 /26 Red-black tree properties Every node in a red-black tree is either black or red Every null leaf is black No path from a leaf to a root can have two.
1 Binary Search Trees Implementing Balancing Operations –AVL Trees –Red/Black Trees Reading: 11.5-11.6.

1 Binary Search Trees Implementing Balancing Operations –AVL Trees –Red/Black Trees Reading:
1 2-3-4 Trees and Red-Black Trees Gordon College Prof. Brinton.

Trees and Red-Black Trees Gordon College Prof. Brinton.
Red Black Trees Colored Nodes Definition Binary search tree.

Red Black Trees Colored Nodes Definition Binary search tree.
CS2420: Lecture 31 Vladimir Kulyukin Computer Science Department Utah State University.

CS2420: Lecture 31 Vladimir Kulyukin Computer Science Department Utah State University.
Multi-Way search Trees 1.2-3 Trees: a. Nodes may contain 1 or 2 items. b. A node with k items has k + 1 children c. All leaves are on same level.

Multi-Way search Trees Trees: a. Nodes may contain 1 or 2 items. b. A node with k items has k + 1 children c. All leaves are on same level.
1 Red Black Trees (Guibas Sedgewick 78). 2 Goal Keep sorted lists subject to the following operations: find(x,L) insert(x,L) delete(x,L) catenate(L1,L2)

1 Red Black Trees (Guibas Sedgewick 78). 2 Goal Keep sorted lists subject to the following operations: find(x,L) insert(x,L) delete(x,L) catenate(L1,L2)
Bottom-Up Splay Trees–Analysis Actual and amortized complexity of join is O(1). Amortized complexity of search, insert, delete, and split is O(log n).

Bottom-Up Splay Trees–Analysis Actual and amortized complexity of join is O(1). Amortized complexity of search, insert, delete, and split is O(log n).
B-Trees (continued) Analysis of worst-case and average number of disk accesses for an insert. Delete and analysis. Structure for B-tree node.

B-Trees (continued) Analysis of worst-case and average number of disk accesses for an insert. Delete and analysis. Structure for B-tree node.
B-Trees Large degree B-trees used to represent very large dictionaries that reside on disk. Smaller degree B-trees used for internal-memory dictionaries.

B-Trees Large degree B-trees used to represent very large dictionaries that reside on disk. Smaller degree B-trees used for internal-memory dictionaries.
Balanced Binary Search Trees height is O(log n), where n is the number of elements in the tree AVL (Adelson-Velsky and Landis) trees red-black trees get,

Balanced Binary Search Trees height is O(log n), where n is the number of elements in the tree AVL (Adelson-Velsky and Landis) trees red-black trees get,

Similar presentations

Ads by Google