4. Trees. 4.1 Preliminaries. 4.2 Binary trees. 4.3 Binary search trees. 4.4 AVL trees. 4.5 Splay trees. 4.6 B-trees. 4. Trees
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1 4. Trees 4.1 Preliminaries 4.2 Binary trees 4.3 Binary search trees 4.4 AVL trees 4.5 Splay trees 4.6 B-trees Malek Mouhoub, CS340 Fall
2 4.1 Preliminaries A Root B C D E F G Height=3 Leaves H I J K L M N P Q Malek Mouhoub, CS340 Fall
3 Terms Child Parent Sibling : nodes that share a common parent. Leaf : a node with no children. Node depth : the number of links on the path from the root to the node. Tree height : the depth of the deepest node. Malek Mouhoub, CS340 Fall
4 Recursive view A tree is either : Empty Contains a root and N s (N 0). Malek Mouhoub, CS340 Fall
5 Implementation of Trees The tree is a collection of nodes : struct TreeNode g; Object element; TreeNode ΛfirstChild; TreeNode ΛnextSibling; f The tree stores a reference to the root node, which is the starting point. Malek Mouhoub, CS340 Fall
6 Implementation of Trees A Root B C D E F G Height=3 H I J K L M N P Q Malek Mouhoub, CS340 Fall
7 Tree Traversals // List a directory in a hierarchical file system void FileSystem::listAll(int depth = 0) const f printname( depth ); if (isdirectory()) for each file c in this directory (for each child) c.listall(depth + 1); g; // Calculate the size of a directory void FileSystem::size( ) const f int totalsize = sizeofthisfile(); if (isdirectory()) for each file c in this directory (for each child) totalsize += c.size(); return totalsize; g; Malek Mouhoub, CS340 Fall
8 4.2 Binary Trees Λ Recursive view A binary tree is either : Empty Contains a root and N binary s (0» N» 2). Malek Mouhoub, CS340 Fall
9 Implementation struct BinaryNode g; Object element; // the data in the node BinaryNode Λleft; // left child BinaryNode Λright; // right child f Malek Mouhoub, CS340 Fall
10 Expression trees Inorder traversal ) infix notation. postorder traversal ) postfix notation. preorder traversal ) prefix notation. + + * a * + g b c * f d e Malek Mouhoub, CS340 Fall
11 Constructing an expression tree Converting a postfix expression into an expression tree : Same principle as evaluating an expression in postfix notation. Use a stack of pointers. For infix expression : Step 1 : Use the algorithm seen in chapter 3 to convert the expression in postfix notation. Step 2 : Use the algorithm above to produce the corresponding expression tree. Malek Mouhoub, CS340 Fall
12 4.3 Binary Search Trees : The Binary Search Tree ADT Goals : Binary search supports Find in O(log N ) worst-case time, but Insert and Remove O(N are ). Would like to support all three operations in O(log N ) worst-case time. Today s result : can support all three operations in O(log N ) average-case time. Later : do it in O(log N ) worst-case time. Basic Ideas : Use a tree to show the logic applied by a binary search. Malek Mouhoub, CS340 Fall
13 Binary Search Expanded Malek Mouhoub, CS340 Fall
14 Binary Search Tree Order Ordering property : for every node in the tree, all items in left are smaller and all items in right are larger (assume no duplicates). X < X > X Malek Mouhoub, CS340 Fall
15 Thinking Recursively Computing the height of a tree is complex without recursion. The height of a tree is one more than the maximum of the heights of the s. Hl+1 X Hr+1 Hl < X > X Hr Malek Mouhoub, CS340 Fall
16 Routine to Compute Height Handle base case (empty tree). Use previous observation for other case. int height(binarynode ΛT) f If (T == null) return -1; else return 1 + Max(height(T->left),height(T->right)); g Malek Mouhoub, CS340 Fall
17 ADT Binary Search Tree Structure BinarySearchTree is Data : A set of N nodes (N 0). Functions : for every bstree 2 Binary Search Tree, item 2 Comparable BynarySearchTree Create(item) ::= return a Binary Search Tree containing item Comparable findmax(bstree) ::= return the maximum of bstree Comparable findmin(bstree) ::= return the minumum of bstree Comparable find(bstree,item) ::= return the matching element if found and ITEM NOT FOUND otherwise. Bool isempty(list) ::= return bstree == empty BinarySearchTree insert(bstree,item) ::= add item to bstree and return the new tree BinarySearchTree remove(bstree,item) ::= removes item and return the new tree Malek Mouhoub, CS340 Fall
18 Implementation template Comparable> <class class BinarySearchTree; template Comparable> <class class BinaryNode f Comparable element; BinaryNode *left; BinaryNode *right; BinaryNode( const Comparable & theelement, BinaryNode *lt, BinaryNode *rt ) : element( theelement ), left( lt ), right( rt ) f g 10 friend class BinarySearchTree<Comparable>; g; Malek Mouhoub, CS340 Fall
19 template Comparable> <class class BinarySearchTree f public: explicit BinarySearchTree( const Comparable & notfound ); BinarySearchTree( const BinarySearchTree & rhs ); BinarySearchTree( ); const Comparable & findmin( ) const; const Comparable & findmax( ) const; 10 const Comparable & find( const Comparable & x ) const; bool isempty( ) const; void printtree( ) const; void makeempty( ); void insert( const Comparable & x ); void remove( const Comparable & x ); const BinarySearchTree & operator=( const BinarySearchTree & rhs ); Malek Mouhoub, CS340 Fall
20 private: BinaryNode<Comparable> *root; const Comparable ITEM NOT FOUND; const Comparable & elementat( BinaryNode<Comparable> *t ) const; void insert( const Comparable & x, BinaryNode<Comparable> * & t ) const; void remove( const Comparable & x, BinaryNode<Comparable> * & t ) const; BinaryNode<Comparable> * findmin( BinaryNode<Comparable> *t ) const; BinaryNode<Comparable> * findmax( BinaryNode<Comparable> *t ) const; BinaryNode<Comparable> * find( const Comparable & x, BinaryNode<Comparable> *t ) const; void makeempty( BinaryNode<Comparable> * & t ) const; 10 void printtree( BinaryNode<Comparable> *t ) const; * clone( BinaryNode<Comparable> *t ) const; BinaryNode<Comparable> g; Malek Mouhoub, CS340 Fall
21 Searching : the find function Set the current node to the root. Repeat : If current node is null, then item is not found. If item is smaller than what is stored in the current node, then branch left. If item is larger than what is stored in the current node, than branch right. Otherwise. we have a match. Malek Mouhoub, CS340 Fall
22 findmin and findmax For findmin, repeatedly branch left until you reach a node with no left child. For findmax, repeatedly branch right until you reach a node with no right child. Malek Mouhoub, CS340 Fall
23 Running Time Time to perform a search is proportional to the depth of the node that terminates the search. For the perfectly balanced tree, this is roughly log N. Malek Mouhoub, CS340 Fall
24 Insertion A new item can be inserted by placing it at the location where an unsuccessful search for it terminates new item 18 Malek Mouhoub, CS340 Fall
25 Coding Simplest code uses recursion.four cases : Case 1 : If the tree is empty, create and return a new node tree. Case 2 : If the item matches the item in the root node do nothing (no duplicates allowed). Case 3 : If the item is less than the item in the root node, recursively insert it in left. Case 4 : If item is greater than the item in the root node, recursively insert in right. Malek Mouhoub, CS340 Fall
26 template <class Comparable> void BinarySearchTree<Comparable>:: insert( const Comparable & x, BinaryNode<Comparable> * & t ) const if( t == NULL ) f t = new BinaryNode<Comparable>( x, NUL L, NULL ); else if( < x t >element ) insert( x, t >left ); else if( < t >element x ) insert( x, t >right ); 10 else ; // Duplicate; do nothing g Malek Mouhoub, CS340 Fall
27 Running Time Running time is still proportional to depth. Arbitrary insertion no longer guarantees that the depth of the tree O(log N is ). However it can be shown that the depth of the tree will be O(log N ) on the average, where average means all possible insertion sequences are equally likely. Malek Mouhoub, CS340 Fall
28 Worst case The most common insertion sequences produce the worst tree. Inserting items in sorted order is disastrous, O(N ) yielding operations. A B C D E Malek Mouhoub, CS340 Fall
29 Bottom Line Insertion and access will be logarithmic on average, but this is meaningful only if insertion sequence is reasonably random. If insertion sequence has some order in it, the binary search tree will degenerate. Need methods of maintaining balance in the tree that preserve a logarithmic worst-case bound. In the following, we will examine two candidates : AVL tree and splay tree. Malek Mouhoub, CS340 Fall
30 Deletion Deletion algorithm is complicated. Basic problem : if the node is deleted, it potentially disconnects the tree. Standard algorithm breaks into three cases : Node to be deleted is a leaf. Node to be deleted has one child. Node to be deleted has two children. Malek Mouhoub, CS340 Fall
31 Deletion of a Leaf Simplest case, because leaf does not connect two parts of the tree. Wipe out the leaf node (its parent should reference a null TreeNode instead of the leaf). Malek Mouhoub, CS340 Fall
32 Deletion of a 1 child node Bypass the node : parent of deleted node gets new child (example below deletes 15) delete item Note : root with one child is a special case, because it does not have a parent. In this case, we simply obtain a new root. Malek Mouhoub, CS340 Fall
33 Deletion of two child node Step 1 : replace node contents with smallest key in right. Example below deletes delete ?? Malek Mouhoub, CS340 Fall
34 Deletion of two child node Step 2 : delete node used as replacement ?? The second step in the two-child case is guaranteed to be simple because that node cannot have a left child (why???). Malek Mouhoub, CS340 Fall
35 Implementation Simple implementation is recursive. Some cases can be combined. The deletion algorithm potentially destroys the balance of the tree because it is a biased algorithm. Notice that in the two child case, it always reduces the right s size, potentially making the left tree heavy after a large number of deletion/insertion pairs. In practice : not noticeable at all. Malek Mouhoub, CS340 Fall
36 template <class Comparable> void BinarySearchTree<Comparable>:: remove( const Comparable & x, BinaryNode<Comparable> * & t ) const f if( t == NULL ) return; // Item not found; do nothing if( x < t >element ) remove( x, t >left ); else if( t >element < x ) remove( x, t >right ); 10 else if( t >left!= NULL && t >right!= NULL ) // Two childre f t >element = findmin( t >right ) >element; g remove( t >element, t >right ); else f BinaryNode<Comparable> *oldnode = t; t = ( t >left!= NULL )? t >left : t >right; delete oldnode;gg 20 Malek Mouhoub, CS340 Fall
37 Printing contents of the tree Easy to print entire tree in sorted order. Case 1 : if tree is empty, print nothing. Case 2 : if tree is not empty : Recursively print left tree. Print item at the root, Recursively print right tree. Strategy is an inorder traversal. Running time is O(N ) : constant work per item. Malek Mouhoub, CS340 Fall
38 4.4 AVL Trees The structure of the tree depends on the order in which keys are entered. Entering the letters A-G in alphabetical order will produce a degenerate tree that is nothing more than a linked list. The solution of this problem is somehow to reorganize the nodes of the tree as we receive new keys, maintaining a near optimal tree structure. AVL trees are used for handling such organization. An AVL tree is a height-balanced 1-tree or HB(1) tree : the maximum difference allowed between the heights of any two s sharing a common root, is one. The two features that make AVL trees important are : by setting a maximum allowable difference in the height of any two s, AVL trees guarantee a minimum level of performance in searching; and Maintaining a tree in AVL form as new nodes are inserted involves the use of one of a set of four possible rotations. Malek Mouhoub, CS340 Fall
39 AVL trees AVL technique used to keep a tree in height-balance : Each time an insertion is made, one must do the following : 1. Let the node to be inserted travel down the appropriate branch, keeping track along the way of the deepest level node on that branch that has a balance factor of +1 or -1 (this particular node is called pivot). 2. Inclusive and below the pivot node, recompute all balance factors along the insertion path traced in step Determine whether the absolute value of the pivot node s balance factor switched from 1 to If there was such a switch, perform a manipulation of tree pointers centered at the pivot node to bring the tree back into height-balance. This operation is frequently referred to as an AVL-rotation. Malek Mouhoub, CS340 Fall
40 AVL rotations 4 cases to consider : Case 1 : The insertion that unbalanced the tree occurred in the left of the left child of the pivot node. Case 2 : The insertion that unbalanced the tree occurred in the right of the right child of the pivot node. Case 3 : The insertion causing the imbalance occurred in the right of the left child of the pivot node. Subcase 1 : Neither the pivot node nor its left child has a right child. Subcase 2 : Insertion is in the left of the right child of the left child of the pivot. Subcase 3 : Insertion occurs in the right of the right child of the left child of the pivot. Case 4 : The insertion that causes the imbalance in the tree is made in the left of the right child of the pivot node. Case 4 is to case 3 as case 2 is to case 1. Malek Mouhoub, CS340 Fall
41 Case 1 Pivot S BF=+1 rotation Pivot S BF=+2 Height = h+2 BF=0 Left of M M Right of M Right of S Height = h inserting a new key BF=+1 Left of M M Right of M Right of S insert balancing Pivot M BF=0 Height = h+2 Left of M insert Old right of M S BF=0 Right of S Malek Mouhoub, CS340 Fall
42 Case 2 Pivot S BF=-1 rotation Pivot S BF=-2 Height = h+2 Height = h Left of S BF=0 Left of V V Right of V inserting a new key Left of S BF=-1 Left of V V Right of V insert balancing Pivot V BF=0 Height = h+2 BF=0 Left of S S Old left of V Right of V insert Malek Mouhoub, CS340 Fall
43 Case 3 : subcase 1 Pivot S BF=1 Pivot S BF=2 Pivot Q BF=0 M BF=0 inserting a new key M BF=-1 balancing M BF=0 S BF=0 Q BF=0 Malek Mouhoub, CS340 Fall
44 Case 3 : subcase 2 Pivot S BF=+1 Pivot S BF=+2 Height = h+2 BF=0 M BF=0 Q Right of S Height = h inserting a new key BF=-1 M BF=1 Q Right of S Left of M Height = h Left of Q Right of Q Height = h-1 Left of M Left of Q insert Right of Q Balancing Pivot Q BF=0 Height = h+2 BF=0 Left of M M Old left of Q insert Old right of Q S BF=-1 Right of S Malek Mouhoub, CS340 Fall
45 Case 3 : subcase 3 Pivot S BF=+1 Pivot S BF=+2 Height = h+2 BF=0 M BF=0 Q Right of S Height = h inserting a new key BF=-1 M BF=-1 Q Right of S Left of M Height = h Left of Q Right of Q Height = h-1 Left of M Left of Q Right of Q insert Balancing Pivot Q BF=0 Height = h+2 BF=-1 Left of M M Old left of Q Old right of Q insert S BF=0 Right of S Malek Mouhoub, CS340 Fall
46 4.5 Splay Trees Think in terms of a set of operations instead of a single one. Goal : guarantees that any M consecutive tree operations starting from an empty tree take at O(M log N ) most time. Idea : After a node is accessed, it is pushed to the root by a series of AVL tree rotations. Malek Mouhoub, CS340 Fall
47 the wrong way Perform single rotations, bottom up. In the case of inserting the keys 1; 2; : : : ; N into an empty tree : O(N ) is required to build the tree. Ω(N 2 ) for accessing all the keys in order. After the keys are accessed, the tree reverts to its original state. Malek Mouhoub, CS340 Fall
48 Splaying If the parent of the node to be pushed is the root, perform a single rotation. Otherwise perform a double rotation : 2 cases to consider (plus symmetries). Malek Mouhoub, CS340 Fall
49 Case 1 G X P X D P G A B C D B C Malek Mouhoub, CS340 Fall
50 Case 2 G X X P C D A B P G B C D Malek Mouhoub, CS340 Fall
51 Advantage of the method When inserting items 1; 2; 3; : : : ; N into an initially empty tree : O(N ) is required to insert all items (same as 1st method). After accessing and pushing node 1 to to root, N=2 units only are required to access node 2 (instead of N 2 for the 1s method). After accessing and pushing the other nodes, the depth becomes log N. Malek Mouhoub, CS340 Fall
52 4.6 B-Trees Problem : Disk utilization of a binary search tree is extremely inefficient. A minimum of a single page is read (at least 512 bytes) in order to get the following information : key value, address of left and right s. First Solution : Divide a binary tree into pages and then store each page in a block of contiguous locations on disk. The number of seeks associated to any search will be then reduced. Problem when inserting new items. ) use B-trees. Malek Mouhoub, CS340 Fall
53 Paged Binary Trees Malek Mouhoub, CS340 Fall
54 B-Trees A B-tree of order M is an M-ary tree with the following properties : 1. The data items are stored at leaves. 2. The non leaf nodes store up to M-1 keys to guide the searching; key i represents the smallest key in i The root is either a leaf or has between two and M children. 4. All non leaf nodes (except the root) have between [M/2] and M children. 5. All leaves are at the same depth and have between [L/2] and L children, for some L. Malek Mouhoub, CS340 Fall
55 B-Tree of order Malek Mouhoub, CS340 Fall
56 Calculating M, L and the level of the B-Tree Example : driving records for citizens in the state of Florida 10,000,000 records. Record (data) : 256 bytes. Key (representing a name) : 32 bytes. Pointer (branch) : 4 bytes (address of disk block). Node = disk block = 8,192 bytes. ) L= 8; = 32 ) (M-1) keys + M pointers = 8,192 ) 32 (M-1) + 4 M = 8,192 ) 36M - 32 = 8,192 ) M ß 228 ) 625,000 leaves. Why? ) Leaves would be on level 4 in the worst case. Why? ) Worst case number of access : O(log M=2 N ). Why? Malek Mouhoub, CS340 Fall
57 Inserting item inserting item 57 Malek Mouhoub, CS340 Fall
58 Inserting 55 : Split into 2 leaves Split into 2 leaves after inserting 55 Malek Mouhoub, CS340 Fall
59 Inserting 40 : split into 2 leaves + split of the parent nserting item 40 causes a split into 2 leaves and then a split of the parent node Malek Mouhoub, CS340 Fall
60 Deleting Malek Mouhoub, CS340 Fall
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