Red-Black Trees (12.2)

Transcription

1 Red-Black Trees v z 02/27/ Red-Black Trees (12.2) A red-black tree can also be defined as a binary search tree that satisfies the following properties: Every node is colored either red or black. (1) The root is black. (2) The children of a red node are black (but the children of a black node may be black or red). (3) All the leaves have the same black depth /27/

2 Black Depth Assign a weight of 1 to each black node a weight of 0 to each red node Define the weight of a leaf the total weight of all the nodes on the path from the root to that leaf. All the leaves must have the same weight. Why red-black trees? BSTs may be unbalanced running time = O(?) The rules of red-black trees ensure that the trees are (almost) balanced running time = O(logN) Longest path 2 x shortest path 02/27/ Height of a Red-Black Tree Theorem: A red-black tree storing n entries has height O(log n) The search algorithm for a red-black tree is the same as that for a binary search tree By the above theorem, searching in a red-black tree takes O(log n) time Demo: /27/

3 Insertion Execute the insertion algorithm for binary search trees and color red the newly inserted node v (unless v is the root, in which case we color it black). If the parent p of v is black, no violations done. Example: inserting 7, 12, 21 02/27/ Insertion Double Red If the parent p of v is red, we have a double red (violation of rule (2)) re-organize the tree by examining the uncle u of v (the parent s sibling). Uncle u is black: this case can t happen (why?) Uncle u is red: recoloring Uncle u is a null pointer: restructuring 02/27/

4 Insertion - Recoloring New node v, its parent p and uncle u are red. Keep v red. Change parent p and uncle u from red to black violating rule (3), so Change parent g of p and u from black to red (unless g is the root, in which case keep it black). 02/27/ Insertion Recoloring (cont d) If the parent of g is red, we get a new double red violation. So we need to go up the tree to reorganize it (percolation) until there is no violation or the root is reached (will be discussed later). 02/27/

5 Insertion - Restructuring New node v and its parent p are red; uncle u is null. There are four restructuring configurations: 02/27/ Insertion Simple Cases Execute the insertion algorithm for binary search trees and color red the newly inserted node v (unless v is the root, in which case we color it black). If the parent p of v is black, no violations done. If the parent p of v is red, we have a double red re-organize the tree by examining the uncle u of v. Uncle u is red: recoloring (may cause a new double red) Uncle u is a null pointer: restructuring Running time of insertion: O(logn) 02/27/

6 Double Red One Level Up On slide 8, it may happen that the parent of g is red a new double red violation. Similar to the simple cases, we examine the uncle of node g The uncle is null: this case can t happen (why?) The uncle is red: recoloring as before. After the recoloring, we may have another double-red repeat the procedure on this slide. The uncle is black: restructuring (rotation). After this rotation, the insertion algorithm terminates. 02/27/ Rotations There are 4 cases (see the next slide) Another restructuring method: cut/link restructure Advantage: no case analysis, more elegant than the 4 rotation cases Disadvantage: there can be more code to write Same time complexity as the 4 rotation cases When to use which method? 4 cases of rotation: writing code cut/link: drawing trees (e.g., for verification during testing) 02/27/

7 02/27/ Cut/Link Procedure The cut/link procedure involves the two "double-red" nodes, which we call the red child and the red parent; the parent of the red parent, which we call the grand-parent (which must be black); the 4 subtrees connected to these 3 nodes (why 4?) for a total of 7 components. We cut the subtree rooted at the grand-parent into 7 components C1, C2,..., C7 so that C1 < C2 <...< C7. Before and after the rotation, the 7 components must be arranged so that their indexes satisfy the property of BSTs: left subtree < current node < right subtree. 02/27/

8 Naming the Components Sort the values of the red child, red parent and grand-parent and assign: smallest node: C2 middle node: C4 largest node: C6 Note: the above assignment and the corresponding indexes satisfy the BST property. Then name the 4 subtrees with C1, C3, C5 and C7 so that they, together with components C2, C4 and C6, satisfy the BST property. 02/27/ Restructuring Arrange and re-color nodes C2, C4 and C6 as shown on slide 9: C4 is the new root and black C2 and C6 are the left and right children of C4 respectively and both red Connect the 4 subtrees C1, C3, C5 and C7 to C2 and C6 as follows: C1 and C3 are the left and right children of C2 respectively. C5 and C7 are the left and right children of C6 respectively. Notes: The above placement preserves the BST property. The colors of the nodes in subtrees C1, C3, C5 and C7 are unchanged. 02/27/

9 Next Topic Hashing (Chapter 5) 02/27/