1.
Balanced Trees Solution• Binary search trees: if all levels filled, then • Keep the trees height balanced (for every search, insertion and deletion are O(log N). node, the difference in height between left• However, performance may deteriorate to and right subtrees at most 1) linear if nodes are inserted in order: • Performance always logarithmic. 10 11 12 Example How to keep the tree balancedLevel 0 10 • Still need to insert preserving the order of 7Level 1 13 keys (otherwise search does not work)Level 2 • Insert as the new items arrive (cannot wait 6 12 15 till have all items)Level 3 • So need to alter the tree as more items 14 arrive. Top down and bottom up Examples insertion• Top down insertion algorithms make • Bottom up: AVL trees changes to the tree (necessary to keep the • Top down: red-black trees tree balanced) as they search for the place to insert the item. They make one pass through the tree. • Exposition (not in Shaffer): follows• Bottom up: first insert the item, and then M.A.Weiss, Data structures and problem work back through the tree making changes. solving using Java, 2nd edition. Less efficient because make two passes through the tree. 1
2.
AVL Trees Non-example and example• AVL (Adelson-Velskii & Landis) trees are Not an AVL: AVL: binary search trees where nodes also have X (2) B additional information: the difference in depth between their left and right subtrees (balance factor). First example of balanced A B (1) X D trees. C D A C E• It is represented as a number equal to the depth of the right subtree minus the depth of E the left subtree. Can be 0,1 or -1. Depth of an AVL tree Insertion in AVL Trees• Calculating the maximal depth of an AVL • Insert new leaf node, as for ordinary binary tree is more difficult than calculating it for search tree. prefectly balanced or complete binary trees. • Then work back up from new leaf to root,• An (almost complete) calculation is given in checking if any height imbalance has been Weiss’s book (Ch.19.4.1). It gives introduced (computing new balance approximately 1.44 log(N+2) - 1.328 as the factors). upper bound on the height, so it’s O(log N). • Perform rotation to correct height imbalance• An average depth of a node in an AVL tree (which rotation depends on the balance is not established analytically yet. factor). Rotations Right rotation• The restructuring is done by performing a • Right rotation around X (tree is heavier on rotation. the left):• A rotation is performed around some node X (-2) X. It can be • single right rotation, A • single left rotation or B C • a double rotation. 2
3.
Right rotation Right rotation• Right rotation around X: • X moves down and to the right, into the position of its right child; X (-2) A • Xs right subtree is unchanged. • Xs left child A moves up to take Xs place. A B X X is now As new right child. • This leaves the old right child of A B C C unaccounted for. Since it comes from the left of X, it is less than X. So it becomes Xs new left child. Left rotation Left rotation• The tree is heavier on the right: • Left rotation around X: X (2) X (2) B A B A B X D C D C D A C E E E Double rotations Example• Two single rotations in opposite directions • Rotating right around 30 will not help: (around two different nodes). 20 (2)• Needed when there is a bend in the branch: 30 10 25 (2) 25 30 27 27 3
4.
Example Example• First single left rotation around 25: • First single left rotation around 25: 20 (2) 20 (2) 10 30 (-2) 10 30 (-2) 25(1) 27 (-1) 27 25 Example Which rotation?• Then single right rotation around 30: • Balance factor -2 (left subtree longer), 20 (1) balance factor of the left daughter -1 or 0: single right rotation. 10 27 • Balance factor 2 (right subtree longer), balance factor of the right daughter 1 or 0: 25 30 single left rotation Which rotation? Deleting in AVL Trees• Balance factor -2, balance factor of the left • Find leftmost node/subtree of deleted nodes daughter 1: double rotation, around the right subtree. daughter to the left, around the root to the • Delete leftmost subtree. right. • Work back up to deleted node, restoring any• Balance factor 2, balance factor of the right imbalances. daughter -1: double rotation, around the • Replace deleted node by leftmost subtree. daughter to the right, around the root to the • Work back up to root, restoring any left. imbalances. 4
5.
Number of rotations Implementation?• Insertion: balance can be restored with a • Not really worth implementing, since not single rotation. the most efficient balanced tree.• Deletion: may require multiple rotations • Have some pseudocode for computing (sometimes nodes are just marked as balance factors in deleted: lazy deletion). www.cs.nott.ac.uk/~nza/G5BADS00/lecture13.html• Overall performance: O(log N).• On average, better than binary search trees. Red-Black Trees Example (black=grey, red=white) 30• Red-black trees are binary search trees which come with additional information 15 70 about the nodes: the nodes are coloured red or black. 20 60 85 • The root is always black; 10 • A red node cannot have red children; • Every path from the root to a null node (a leaf 5 50 65 80 90 or a node with only one child) should have the same number of black nodes. 3 40 55 Non-example Depth of Red-Black Trees 30 • The depth of a red-black tree is at most 15 70 2 log(N+1) (this can be shown by induction). • Intuitively, since we can pad some branches 10 20 60 85 with red nodes and compose other branches only of black ones, you expect the increase 5 50 65 80 90 in depth compared to perfectly balanced trees (about twice longer…) 40 55 5
6.
Red-Black Trees insertion The ideaInsertion can be done top down, changing • If we need to insert an item in the red-black colours and rotating subtrees while tree, and we arrive at the insertion point searching for a place to insert the node. The which is a node S which does not have at rules should be satisfied after the new node least one of the daughters and S is black we is inserted. Only requires one pass through are done: we insert our item as a red node the tree and is guaranteed to be O(log N). and nothing changes. • So make sure that whatever colour S was before, when we arrive there it’s black. Example: insert 91 Example: insert 91 30 30 15 70 15 70 10 20 60 85 10 20 60 85 5 50 65 80 90 5 50 65 80 90 40 55 40 55 91 Insertion Example: single rotation 1• Insert always as a red node• If the parent in black, we are done 20 18• If the parent is red, could perform a rotation 18 22 15 20 with a colour change (see below) 15 19 21 23 10 16 19 22• If the sibling of the parent node is black, this rotation solves the problem 21 23 10 16• If not, we are stuck, so we need to remove a possibility of two red siblings. 6
7.
Example: single rotation 2 (problem) Top down insertion • On the way down, if we see a node X which 20 18 has two red children, make X red and 15 children black (if X is root, X will be 18 22 20 recoloured black). 10 16 19 15 19 21 23 22 • This preserves the number of black nodes on the path, but may not preserve the 21 23 10 16 property that there are no consecutive red nodes on any path. Example of a colour flip Top down insertion contd. • Fix consecutive red nodes: perform a 30 30 rotation with a colour change. • As in AVL trees, different kinds of rotations 15 70 15 70 depending on the kind of problem. Also involve colour change. 60 85 60 85 Java implementation Summary• See Weiss, Chapter 19.5.3 (insertion only). • Binary search trees can become imbalanced, leading to inefficient search• Code available on • AVL trees and red-black trees: heighthttp://www.cs.fiu.edu/~weiss/dsaajava/Code/ balanced binary search trees. • Leads to reasonable search complexity: O(log N). • Efficient (but complicated) algorithms for maintaing height balance under insertion and deletion. • Requires rotations and colour flips to restore balance. 7
Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.
Be the first to comment