Red-Black Trees: Data Structure and Algorithms, Study notes of Data Representation and Algorithm Design

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Red-Black Trees

Based on materials by Dennis Frey, Yun Peng, Jian Chen, and Daniel Hood

Advanced Data Structures

n CS 206 covered basic data structures q Lists, binary search trees, heaps, hash tables n CS 246 will introduce you to some advanced data structures and their use in applications q Red-Black Trees: a type of self-balancing BST q KD-Trees: a type of space partitioning tree q Graphs: represents a set of entities and relations n Over the next few weeks, we will discuss these data structures, starting today with Red-Black Trees

Review of Tree Rotations: Zig-Zig

(Node and Parent are Same Side)

Rotate P around G, then X around P

Review of Tree Rotations: Zig-Zag

(Node and Parent are Different Sides)

Rotate X around P, then X around G

Red-Black Trees

n Definition: A red-black tree is a binary search tree in which: q Every node is colored either Red or Black. q Each NULL pointer is considered to be a Black “node”. q If a node is Red, then both of its children are Black. q Every path from a node to a NULL contains the same number of Black nodes. q By convention, the root is Black n Definition: The black-height of a node X in a red-black tree is the number of Black nodes on any path to a NULL, not counting X.

A Red-Black Tree with NULLs shown Black-Height of the tree (the root) = 3 Black-Height of node “X” = 2 X

Black Height of the tree? Black Height of X? X

Theorem 1 – Any red-black tree with root x , has n ≥ 2 bh(x)

- 1 nodes, where bh(x) is the black height of node x. Proof: by induction on height of x.

Theorem 3 – In a red-black tree, no path from any node, X, to a NULL is more than twice as long as any other path from X to any other NULL. Proof: By definition, every path from a node to any NULL contains the same number of Black nodes. By Theorem 2, a least ½ the nodes on any such path are Black. Therefore, there can no more than twice as many nodes on any path from X to a NULL as on any other path. Therefore the length of every path is no more than twice as long as any other path.

Theorem 4 – A red-black tree with n nodes has height h ≤ 2 lg( n + 1). Proof: Let h be the height of the red-black tree with root x. By Theorem 2, bh(x) ≥ h/ From Theorem 1, n ≥ 2 bh(x)

• 1 Therefore n ≥ 2 h/
• 1 n + 1 ≥ 2 h/ lg(n + 1) ≥ h/ 2lg(n + 1) ≥ h

Bottom –Up Insertion

n Insert node as usual in BST n Color the node Red n What Red-Black property may be violated? q Every node is Red or Black? q NULLs are Black? q If node is Red, both children must be Black? q Every path from node to descendant NULL must contain the same number of Blacks?

Bottom Up Insertion

n Insert node; Color it Red; X is pointer to it n Cases 0: X is the root -- color it Black 1: Both parent and uncle are Red -- color parent and uncle Black, color grandparent Red. Point X to grandparent and check new situation. 2 (zig-zag): Parent is Red, but uncle is Black. X and its parent are opposite type children -- color grandparent Red, color X Black, rotate left(right) on parent, rotate right(left) on grandparent 3 (zig-zig): Parent is Red, but uncle is Black. X and its parent are both left (right) children -- color parent Black, color grandparent Red, rotate right(left) on grandparent

X

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Case 2 – Zig-Zag Double Rotate X around P; X around G Recolor G and X

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Case 3 – Zig-Zig Single Rotate P around G Recolor P and G