# Chapter 4: Trees Part II - AVL Tree

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Chapter 4: Trees Part II - AVL Tree
Mark Allen Weiss: Data Structures and Algorithm Analysis in Java Chapter 4: Trees Part II - AVL Tree

Better Binary Search Trees
Prevent the degeneration of the BST : A BST can be set up to maintain balance during updating operations (insertions and removals) Types of BST which maintain the optimal performance: splay trees AVL trees Red-Black trees B-trees

AVL Trees Balanced binary search trees
An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1.

Height of an AVL Tree Proposition: The height of an AVL tree T storing n keys is O(log n). Justification: The easiest way to approach this problem is to find n(h): the minimum number of internal nodes of an AVL tree of height h. n(1) = 1 and n(2) = 2 for n ≥ 3, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and the other AVL subtree of height n-2.  n(h) = 1 + n(h-1) + n(h-2) given n(h-1) > n(h-2)  n(h) > 2n(h-2) n(h) > 2n(h-2) n(h) > 4n(h-4) n(h) > 2in(h-2i) pick i = h/2 – 1  n(h) ≥ 2 h/2-1 follow h < 2log n(h) +2  height of an AVL tree is O(log n)

Insertion A binary search tree T is called balanced if for every node v, the height of v’s children differ by at most one. Inserting a node into an AVL tree involves performing an expandExternal(w) on T, which changes the heights of some of the nodes in T. If an insertion causes T to become unbalanced, we travel up the tree from the newly created node until we find the first node x such that its grandparent z is unbalanced node. Since z became unbalanced by an insertion in the subtree rooted at its child y, height(y) = height(sibling(y)) + 2 Need to rebalance...

Insertion: Rebalancing
To rebalance the subtree rooted at z, we must perform a restructuring we rename x, y, and z to a, b, and c based on the order of the nodes in an in-order traversal. z is replaced by b, whose children are now a and c whose children, in turn, consist of the four other subtrees formerly children of x, y, and z.

Insertion (cont’d) unbalanced... ...balanced

Restructuring The four ways to rotate nodes in an AVL tree, graphically represented -Single Rotations:

Restructuring (cont’d)
double rotations:

Restructure Algorithm
Algorithm restructure(x): Input: A node x of a binary search tree T that has both a parent y and a grandparent z Output: Tree T restructured by a rotation (either single or double) involving nodes x, y, and z. 1: Let (a, b, c) be an inorder listing of the nodes x, y, and z, and let (T0, T1, T2, T3) be an inorder listing of the the four subtrees of x, y, and z, not rooted at x, y, or z. Replace the subtree rooted at z with a new subtree rooted at b Let a be the left child of b and let T0, T1 be the left and right subtrees of a, respectively. Let c be the right child of b and let T2, T3 be the left and right subtrees of c, respectively.

Let’s go into a little more detail on this algorithm... Any tree that needs to be balanced can be grouped into 7 parts: x, y, z, and the 4 trees anchored at the children of those nodes (T0-3)

Make a new tree which is balanced and put the 7 parts from the old tree into the new tree so that the numbering is still correct when we do an in-order-traversal of the new tree. This works regardless of how the tree is originally unbalanced. Let’s see how it works!

Number the 7 parts by doing an in-order-traversal. (note that x,y, and z are now renamed based upon their order within the traversal)

Now create an Array, numbered 1 to 7 (the 0th element can be ignored with minimal waste of space) Cut() the 4 T trees and place them in their inorder rank in the array

Now cut x,y, and z in that order (child,parent,grandparent) and place them in their inorder rank in the array. Now we can re-link these subtrees to the main tree. Link in rank 4 (b) where the subtree’s root formerly

Link in ranks 2 (a) and 6 (c) as 4’s children.

Finally, link in ranks 1,3,5, and 7 as the children of 2 and 6. Now you have a balanced tree!

This algorithm for restructuring has the exact same effect as using the four rotation cases discussed earlier. Advantages: no case analysis, more elegant Disadvantage: can be more code to write Same time complexity

Removal We can easily see that performing a removeAboveExternal(w) can cause T to become unbalanced. Let z be the first unbalanced node encountered while traveling up the tree from w. Also, let y be the child of z with the larger height, and let x be the child of y with the larger height. We can perform operation restructure(x) to restore balance at the subtree rooted at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached

Removal (cont’d) example of deletion from an AVL tree: y x T T y z x T
88 44 17 78 32 50 48 62 1 4 2 3 54 T y x T 2 1 88 17 78 50 48 62 2 3 54 T y x 44 4 z

Removal (cont’d) example of deletion from an AVL tree: x y z T 88 17
78 50 48 62 1 4 2 3 54 T y x z 2 44

B-Trees Up to now, all data that has been stored in the tree has been in memory. If data gets too big for main memory, what do we do? If we keep a pointer to the tree in main memory, we could bring in just the nodes that we need. For instance, to do an insert with a BST, if we need the left child, we do a disk access and retrieve the left child. If the left child is NIL, then we can do the insert, and store the child node on the disk. Not too good for a BST

B-Trees The problem with BST: storing the data requires disk accesses, which is expensive, compared to execution of machine instructions. If we can reduce the number of disk accesses, then the procedures run faster. The only way to reduce the number of disk accesses is to increase the number of keys in a node. The BST allows only one key per leaf. Very good and often used for Search Engines! (when collection size gets very big  the index does not fit in memory)

B-Trees If we increase the number of keys in the nodes, how will we do any tree operations effectively? Above is a node with 7 keys. How do we add children?

B-Trees Clearly, the tree below is useless. 10 20 30 40 50 60 70
How many pointers do we need? Using the idea of BST, we need to be able to put nodes into the tree that have smaller, same and larger values than the node we are currently examining.

B-Trees: A General Case of Multi-Way Search Trees
70 60 50 40 30 20 10 8 7 9 6 4 3 2 1 11 22 77 33 44 55 66 We can easily find any value. We need to create operations, which require rules on what makes a tree a B-Tree. Clearly, having one key per node would be very bad. We need a mechanism to increase the height of the tree (since the number of keys in any node can get very high) so we can shift keys out of a node, making the nodes smaller.

B-Trees: Fields in a Node
A B-Tree is a rooted tree (whose root is root[T])) having the following properties: 1. Every internal node x has the following fields: leaf[x] key8[x] key7[x] key6[x] key5[x] key4[x] key3[x] key2[x] key1[x] n[x] c9[x] c8[x] c7[x] c6[x] c5[x] c4[x] c3[x] c2[x] c1[x] n[x] is the number of keys in the node. n[x] = 8 above. leaf[x] = false for internal nodes, since x is not a leaf. The keyi[x] are the values of the keys, where keyi[x]  keyi+1[x]. ci[x] are pointers to child nodes. All the keys in ci[x] have values that are between keyi-1[x] and keyi[x].

B-Trees Leaf nodes have no child pointers
leaf[x] = true for leaf nodes. All leaf nodes are at the same level leaf[x] key8[x] key7[x] key6[x] key5[x] key4[x] key3[x] key2[x] key1[x] n[x]

B-Trees There are lower and upper bounds on the number of keys a node can contain. This depends on the “minimum degree” t  2, which we must specify for any given B-Tree. a. Every node other than the root must have at least t-1 keys. Every internal node other than the root thus has t children. If the tree is nonempty, the root must have at least one key. b. Every node can contain at most 2t-1 keys. Therefore, an internal node can have at most 2t children. A node is full if it contains exactly 2t-1 keys.

Height of B-Tree If n  1, then for any n-key B-tree of height h and mimimum degree t  2, height = h  logt[(n+1)/2] The important thing to notice is that the height of the tree is log base t. So, as t increases, the height, for any number of nodes n, will decrease. Using the formula logax = (logbx)/(logba), we can see that log2106 = (log10106)/(log102)  6/  19 log10106 = 6 So, 13 less disk accesses to get to the leafs!

Basic Operations The root of the B-tree is always in main memory, so that a Disk-Read on the root is never required; a Disk-Write of the root is required, however, whenever the root node is changed. Any nodes that are passed as parameters must already have had a Disk-Read operation performed on them.

Searching a B-Tree Start at the leftmost key
in the node, and go to the right until you go too far. If it is a leaf node, then you are done, as there is no leaf to inspect Otherwise, retrieve the child node from the disk, and put it into memory

Inserting into B-trees
Really very easy. Very similar with (2,4) trees. Just keep in mind that you are starting at the root, and then finding the subtree where the key should be inserted, and following the pointer. A deletion may eventually occur, and sometimes deletions force keys into their parents. So, if we encounter a full node on our way to the node where the insertion will take place, we must split that node into two.

Otherwise, call Nonfull()
Inserting into B-trees (cont’d) B-Tree-Insert If the node has 2t-1 keys, it can’t accept any more keys, so you need to split it into 2 nodes before doing the insert. Otherwise, call Nonfull()

Deleting Keys from Nodes

Deleting Keys from Nodes

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