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Learn about red-black trees, a type of self-balancing binary search tree, that maintain balance and efficient search, insert, and delete operations. Explore the properties, rotations, and examples of insertion and deletion.
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Group Members • Bushra Shabbir SP10-BCS-020 • Mehreen Arif SP10-BCS-040 • Misbah Kiran SP10-BCS-042
Introduction • Binary Tree: is a tree structure in which each node has at most 2 children. • Binary Search Tree: Each node has a value, where the left sub-tree node contains only values lesser or equal to it’s parent, and the right sub-tree contains nodes of greater or equal value to it’s parents.
1 2 3 4 5 Binary Search Tree • In the case of a Binary Search Tree search, insert and delete all have a running time in the order of O (log n) for n items in best case. • A Binary Search tree is not self balancing, so in worst case its running time is O (n). As in case of first 5 integers:
11 2 12 1 7 15 5 8 Balanced Search Tree • Balanced Means: the longest path is no more then twice as long as the shortest path.
Red Black Tree • A red black tree is a type of binary search tree that is self balancing. Red-black trees aim to keep the tree balanced by • Coloring each node in the tree with either red or black • Preserving a set of properties that guarantee that the deepest path in the tree is not longer than twice the shortest one.
Properties • A red-black tree has the following properties: • Every node is colored either Red or Black. • By convention, the root is always Black • Each NULL pointer is considered to be a Black “node”. • If a node is Red, then both of its children are Black. (If not then Red Violation Occurs) • Every path from a node to a NULL contains the same number of Black nodes. (If not then Black Violation occurs)
Height of RBT • Height of a node: • H (x) = number of edges in a longest path to a leaf. • Black-height of a node x, bh (x): • Bh (x)= number of black nodes (including nil)on the path from x to leaf, not counting x. • Black-height of a red-black tree is the black-height of its root.
Rotation • Rotation needs to be talked about before Insert and Delete. • Operations (Tree-Insert and Tree-Delete) can destroy the properties of a red black tree. To restore sometimes pointers in the tree need to be change. This is called “Rotation”. • There are 2 types of rotations, left and right. Note: Sometimes to restore the red black properties, some nodes have to have their colors changed from red to black or black to red.
X Y z Y X V Z W V W Left Rotation Here are the steps involved in for left rotation: • Assume node x is the parent and node y is a non-leaf right child. • Let y be the parent and x be its left child. • Let y’s left child be x’s right child.
Y X z X Y V Z W V W Right Rotation Here are the steps involved in for right rotation: • Assume node x is the parent and node y is a non-leaf left child. • Let y be the parent and x be its right child. • Let y’s right child be x’s left child.
Insertion Insertion can be done in the order of O (log n). It follows closely the insertion into a binary search tree, except • The Node inserted is red • A restore function must be called to fix red-black properties that might be violated.
Example of Inserting Sorted Numbers 1 2 3 4 5 6 7 8 9 10 11 1 1 Insert 1. A leaf so red. Realize it is root so recolor to black. 13
Insert 2 1 2 make 2 red. Parent is black so done. 14
Insert 3 1 2 3 2 1 3 Insert 3. Parent is red. Parent's sibling is black(null) 3 is outside relative to grandparent. Rotate parent and grandparent 15
Insert 4 2 1 3 2 1 3 4 On way down see2 with 2 red children.Recolor 2 red andchildren black.Realize 2 is rootso color back to black When adding 4parent is blackso done. 16
Insert 5 2 1 3 4 5 5's parent is red.Parent's sibling isblack (null). 5 isoutside relative tograndparent (3) so rotateparent and grandparent thenrecolor 17
Finish insert of 5 2 1 4 3 5 18
Insert 6 2 1 4 3 5 On way down see4 with 2 redchildren. Make4 red and childrenblack. 4's parent isblack so no problem. 19
Finishing insert of 6 2 1 4 3 5 6 6's parent is blackso done. 20
Insert 7 2 1 4 3 5 6 7 7's parent is red.Parent's sibling isblack (null). 7 isoutside relative tograndparent (5) so rotate parent and grandparent then recolor 21
Finish insert of 7 2 1 4 3 6 5 7 22
Insert 8 2 1 4 3 6 5 7 On way down see 6with 2 red children.Make 6 red andchildren black. Thiscreates a problembecause 6's parent, 4, isalso red. Must performrotation. 23
Still Inserting 8 2 1 4 3 6 5 7 Re-colored nowneed torotate 24
Finish inserting 8 4 2 3 6 5 7 1 8 Again Re-colored 25
Insert 9 4 2 3 6 5 7 1 8 9 On way down see 4 has two red childrenso recolor 4 red and children black. Realize 4 is the root so recolor black 26
Finish Inserting 9 4 2 3 6 5 8 1 7 9 After rotations and re-coloring 27
Insert 10 4 2 3 6 5 8 1 7 9 10 On way down see 8 has twored children so change 8 tored and children black 28
Insert 11 4 2 3 6 5 8 1 7 9 10 11 Again a rotation isneeded. 29
Finish inserting 11 4 2 3 6 5 8 1 7 10 9 11 30
Deletion • Deletion, like insertion, should preserve all the RBT properties. • The properties that may be violated depends on the color of the deleted node. • Red – OK. Why? • Black?Black Violation • Steps: • Do regular BST deletion. • Fix any violations of RBT properties that may result.