Topic Binary Trees (Non-Linear Data Structures)
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1 Topic Binary Trees (Non-Linear Data Structures) CIS210 1
2 Linear Data Structures Arrays Linked lists Skip lists Self-organizing lists CIS210 2
3 Non-Linear Data Structures Hierarchical representation? Trees General Trees Binary Trees Search Trees Balanced Trees Heaps.. CIS210 3
4 General Trees CIS210 4
5 A (Rooted) Tree? A finite, nonempty set of nodes and edges s.t. One special node (the root of the tree) Each node may be associated (edge) with one or more different nodes (its children). Each node except the root has exactly one parent. The root node has no parent (no incoming edge). There exists a unique path from the root to any other node! CIS210 5
6 General Rooted Trees root... CIS210 6
7 Example: (General Rooted) Trees? CIS210 7
8 More Terminology Path A sequence of nodes. Level of a node Height of a node The number of nodes in the longest path from that node to a leaf. Height of a tree The height of the root node. CIS210 8
9 An N-ary Tree? Each node may be associated (edge) with exactly N different nodes (its children). If the set is empty (no node), then an empty N-ary tree. CIS210 9
10 Example: N-ary Trees (N=3) * Note: = Empty tree! CIS210 10
11 The World of Trees General Trees Exact N children N-ary Trees CIS210 11
12 An Ordered Tree? A rooted tree in which the children of each node are ordered. first child, second child, third child, etc. Most practical implementations of trees define an implicit ordering of the subtrees. CIS210 12
13 Example: Ordered Trees CIS210 13
14 Different Views on Trees We may view trees as A mathematical construct. A data structure. An abstract data type. CIS210 14
15 Binary Trees CIS210 15
16 Binary Trees A binary tree is an ordered N-ary tree where N=2! root T left T right CIS210 16
17 Example: Binary Trees * Note: = Empty tree! CIS210 17
18 Example: Binary Trees? CIS210 18
19 Two Different Binary Trees CIS210 19
20 Quiz How many different binary trees with 3 nodes? 5 How many different binary trees with 4 nodes? 14 CIS210 20
21 Quiz What is the range of possible heights of a binary tree with 3 nodes? 2 to 3 What is the range of possible heights of a binary tree with 100 nodes? 7 to 100 CIS210 21
22 The World of Trees General Trees N-ary Trees N=2 Binary Trees CIS210 22
23 Different Shapes of Binary Trees Short & Fat binary trees A full binary tree A complete binary tree A balanced binary tree Tall & Skinny binary trees A skewed binary tree CIS210 23
24 A Full Binary Tree A binary tree in which All of the leaves are on the same level. (at level h for the binary tree of height h.) Every nonleaf node has exactly two children. CIS210 24
25 Shape of A Full Binary Tree The basic shape of a full binary tree is triangular! CIS210 25
26 Example: Full Binary Trees CIS210 26
27 A Complete Binary Tree A binary that is Either full binary tree Or full through the next-to-last level, with the leaves on the last level as far to the left as possible (filled in from left to right). A binary tree in which all leaf nodes are at level n or n-1, and all leaves at level n are towards the left. A binary tree of height h that is full to level h-1 and has level h filled from left to right. CIS210 27
28 Shape of A Complete Binary Tree The basic shape of a complete binary tree is like CIS210 28
29 Example: Complete Binary Trees? CIS210 29
30 Example: Complete Binary Trees? CIS210 30
31 A (Height) Balanced Binary Tree A binary tree in which the left and right subtrees of any node have heights that differ by at most 1. CIS210 31
32 Example: (Height) Balanced Binary Trees CIS210 32
33 A Skewed Binary Tree A degenerate binary tree A skewed binary tree is expensive to process! CIS210 33
34 Example: Skewed Binary Trees CIS210 34
35 Quiz Is a full binary tree complete? Yes! Is a complete binary tree full? No! Is a full binary tree balanced? Yes! Is a complete binary tree balanced? Yes! CIS210 35
36 Properties of Binary Trees What is the maximum height of a binary tree with n nodes? n The maximum height of a binary tree with n nodes is n. CIS210 36
37 Properties of Binary Trees What is the minimum height of a binary tree with n nodes? log (n+1) where the ceiling of log (n+1) = log (n+1) rounded up. The minimum height of a binary tree with n nodes is log (n+1). (The ceiling of log (n+1) = log (n+1) rounded up.) CIS210 37
38 Properties of Binary Trees The height of a binary tree with n nodes is at least log (n+1) and at most n. CIS210 38
39 Properties of Binary Trees What is the minimum number of nodes that a binary tree of height h can have? h The minimum number of nodes that a binary tree of height h can have is h. CIS210 39
40 Properties of Binary Trees What is the maximum number of nodes that a full binary tree of height h can have? 2 h - 1 The maximum number of nodes that a binary tree of height h can have is 2 h - 1. CIS210 40
41 Properties of Binary Trees Full binary trees and complete binary trees have the minimum height! Skewed (degenerate) binary trees have the maximum height! CIS210 41
42 Representation of Binary Trees CIS210 42
43 How to Represent a Binary Tree? An array-based representation An array-based representation for complete binary trees A pointer-based representation CIS210 43
44 1. An Array-Based Representation Represent a node in the binary tree as a structure A data Two indexes (One for each child) Represent the binary tree as an array of structures. CIS210 44
45 An Array-Based Representation const int MAX_NODES = 100; struct BTnode { typedef string DataType; DataType Data; int Lchild; int Rchild; }; BTnode BT[MAX_NODES]; int Root; int Free; CIS210 45
46 An Array-Based Representation Jane Bob Tom Alan Ellen Nancy BT Data Lchild Rchild Root Jane 1 2 Bob 3 4 Tom 5-1 Alan -1-1 Ellen -1-1 Nancy Free 6 CIS210 46
47 2. An Array-Based Representation of a Complete Binary Tree A better array-based representation for a complete binary tree! Represent a node in the binary tree as A data Represent the binary tree as an array. CIS210 47
48 An Array-Based Representation of a Complete Binary Tree const int MAX_NODES = 100; typedef string DataType; DataType BT[MAX_NODES]; int Root; CIS210 48
49 An Array-Based Representation of a Complete Binary Tree BT Data Jane Bob Tom Alan Ellen Nancy Jane Bob Tom Alan Ellen Nancy Root BT[0] CIS210 49
50 An Array-Based Representation of a Complete Binary Tree Any node BT[ i ] Its left child = BT [ 2 * i + 1 ] Its left child = BT [ 2 * i + 2 ] Its parent = BT [ (i -1) / 2 ] Must maintain it as a complete binary tree! Limited delete! CIS210 50
51 3. A Pointer-Based Representation Represent a node in the binary tree as a structure A data Two pointers (One for each child) Represent the binary tree as a linked structure. CIS210 51
52 A Pointer-Based Representation Root Jane Jane Bob Tom Bob Tom Alan Ellen Nancy Alan Ellen Nancy CIS210 52
53 A Pointer-Based Representation struct BTnode { typedef string DataType; DataType Data; BTnode* LchildPtr; BTnode* RchildPtr; }; BTnode* rootbt; CIS210 53
54 A Pointer-Based Representation using Template template<class DataType> struct BTnode { DataType Data; BTnode<DataType>* LchildPtr; BTnode<DataType>* RchildPtr; }; BTnode<DataType>* rootbt; CIS210 54
55 A Pointer-Based Representation using Template Class template<class DataType> class BTnode { Public: BTnode(); BTnode(DataType D, BTnode<DataType>* l, BTnode<DataType>* r) :data(d), LchildPtr(l), RchildPtr(r) { } friend class BT<DataType>; private: DataType data; BTnode<DataType>* LchildPtr; BTnode<DataType>* RchildPtr; }; CIS210 55
56 A Pointer-Based Representation using Template Class template<class DataType> class BT { Public: BT(); private: BTnode<DataType>* rootbt; }; CIS210 56
57 Binary Trees as ADTs CIS210 57
58 Operations on Binary Trees Create an empty binary tree. Create a one-node binary tree, given an item. Create a binary tree, given an item for its root and two binary trees for the root s left and right subtrees. Attach a left or right child to the binary tree s root. Attach a left or right subtree to the binary tree s root. Detach a left or right subtree to the binary tree s root. Destroy a binary tree. CIS210 58
59 More Operations on Binary Trees... Determine whether a binary tree empty? Determine or change the data in the binary tree s root. Return a copy of the left or right subtree of the binary tree s root. Traverse the nodes in a binary tree in preorder, inorder or postorder. CIS210 59
60 Traversal Operations on Binary Trees It is frequently necessary to examine every node exactly once in a binary tree. Binary tree traversal is the process of Visiting systematically all the nodes of a binary tree and Performing a task (calling a visit procedure like print). CIS210 60
61 Traversal Operations on Binary Trees Two essential approaches: Depth-first traversal Breadth-first traversal CIS210 61
62 Possible Depth-First Traversals Six possible ways to arrange those tasks: 1. Process a node, then left-child subtree, then right-child subtree. 2. Process left-child subtree, then a node, then right-child subtree. 3. Process left-child subtree, then right-child subtree, then a node. 4. Process a node, then right -child subtree, then left-child subtree. 5. Process right -child subtree, then a node, then left-child subtree. 6. Process right -child subtree, then left-child subtree, then a node. In almost all cases, the subtrees are analyzed left to right! CIS210 62
63 Common Binary Tree Traversals Three ways to arrange those tasks: 1. Process a node, then its left-child subtree, then its right-child subtree. Preorder Traversal 2. Process its left-child subtree, then a node, then its right-child subtree. Inorder Traversal 3. Process its left-child subtree, then its right-child subtree, then a node. Postorder Traversal CIS210 63
64 1. Pre-order Traversal If the tree is not empty then Visit the root Preorder traverse the left subtree recursively Preorder traverse the right subtree recursively CIS210 64
65 Pre-order Traversal - Processing Order CIS210 65
66 Example: Preorder traversal A B D A B C D E F G H I C E H F G I CIS210 66
67 Preorder Traversal & Print void preorder_print(btnode* rootbt) { if (rootbt!= NULL) { cout << rootbt->data << endl; preorder_print(rootbt-> LchildPtr); preorder_print(rootbt-> RchildPtr); } } CIS210 67
68 Preorder Traversal and Operation typedef void (*ftype)(datatype& AnItem); void preorder(btnode* rootbt, ftype Op) { if (rootbt!= NULL) { Op(rootBT->Data); preorder(rootbt-> LchildPtr); preorder(rootbt-> RchildPtr); } } CIS210 68
69 2. In-order Traversal If the tree is not empty then Inorder traverse the left subtree recursively Visit the root Inorder traverse the right subtree recursively CIS210 69
70 In-order Traversal - Processing Order CIS210 70
71 Example: Inorder traversal A B D B C A F E G D I H C E H F G I CIS210 71
72 Inorder Traversal & Print void inorder_print(btnode* rootbt) { if (rootbt!= NULL) { inorder_print(rootbt-> LchildPtr); cout << rootbt->data << endl; inorder_print(rootbt-> RchildPtr); } } CIS210 72
73 Inorder Traversal and Operation typedef void (*ftype)(datatype& AnItem); void inorder(btnode* rootbt, ftype Op) { if (rootbt!= NULL) { inorder(rootbt-> LchildPtr); Op(rootBT->Data); inorder(rootbt-> RchildPtr); } } CIS210 73
74 3. Post-order Traversal If the tree is not empty then Postorder traverse the left subtree recursively Postorder traverse the right subtree recursively Visit the root CIS210 74
75 Post-order Traversal - Processing Order CIS210 75
76 Example: Postorder traversal A B D C B F G E I H D A C E H F G I CIS210 76
77 Postorder Traversal & Print void postorder_print(btnode* rootbt) { if (rootbt!= NULL) { postorder_print(rootbt-> LchildPtr); postorder_print(rootbt-> RchildPtr); cout << rootbt->data << endl; } } CIS210 77
78 Postorder Traversal and Operation typedef void (*ftype)(datatype& AnItem); void postorder(btnode* rootbt, ftype Op) { if (rootbt!= NULL) { postorder(rootbt-> LchildPtr); postorder(rootbt-> RchildPtr); Op(rootBT->Data); } } CIS210 78
79 5. Backward In-order Traversal 5. Process right-child subtree, then a node, then leftchild subtree. If the tree is not empty then Backward Inorder traverse the right subtree recursively Visit the root Backward Inorder traverse the left subtree recursively CIS210 79
80 Backward In-order Traversal - Processing Order CIS210 80
81 Example: Backward Inorder Traversal A B D H I D G E F A C B C E H F G I CIS210 81
82 Example: Backward Inorder Traversal B C F A E G D I H H I D G E F A C B A D B H E C I G F CIS210 82
83 Example: Backward Inorder traversal B C F A E G D I H A D B H E C I G F CIS210 83
84 Example: Backward Inorder traversal A B A D B D C F E G I H C E H F G I CIS210 84
85 Breadth-First Traversal of Binary Trees CIS210 85
86 Breadth-First (Levelorder)Traversal If the tree is not empty then then visit the nodes in the order of their level (depth) in the tree. Visit all the nodes at depth zero (the root). Then, all the nodes from left to right at depth one Then, all the nodes from left to right at depth two Then, all the nodes from left to right at depth three And so on... CIS210 86
87 Level-order Traversal - Processing Order CIS210 87
88 Example: Levelorder traversal A B D A B D C E H F G I C E H F G I CIS210 88
89 Expression Tree ADT CIS210 89
90 Algebraic Expressions: Notations Algebraic expressions Fully parenthesized Infix notation Not fully parenthesized Infix notation Postfix notation Prefix notation CIS210 90
91 1. Fully Parenthesized Infix Notation We need to place parentheses around each pair of operands together with their operator! Examples: (1+2) (1+(2 * 3)) ((1 + 2) * 3) ( (a / b) + ( (c - d) * e ) ) Inconvenient! CIS210 91
92 2. Not Fully Parenthesized Infix Notation We can omit unnecessary parentheses! Examples: (1+2) (1+ (2 * 3)) 1+ 2 * 3 ((1 + 2) * 3) (1 + 2) * 3 ( (a / b) + ( (c - d) * e ) ) a / b + (c - d) * e Convenient, BUT, we need rules to interpret correctly. CIS210 92
93 Not Fully Parenthesized Infix Notation Operator precedence rule * / higher than + - Operator association rule Associate from left to right Examples: (Interpretation using rules) (1+2) 1+ 2 * 3 (1+ (2 * 3)) (1 + 2) * 3 ((1 + 2) * 3) a / b + (c - d) * e ( (a / b) + ( (c - d) * e ) ) CIS210 93
94 3. Postfix Notation Postfix Notation: <postfix> := <id> (<postfix> <postfix> <op> ) <op> := + - * / <id> := <variable> <number> Examples: (1+2) (1+ (2 * 3)) 1+ 2 * * + ((1 + 2) * 3) (1 + 2) * * ( (a / b) + ( (c - d) * e ) ) a / b + (c - d) * e a b / c d - e * + CIS210 94
95 4. Prefix Notation Postfix Notation: <prefix> := <id> (<op> <prefix> <prefix>) <op> := + - * / <id> := <variable> <number> Examples: (1+2) (1+ (2 * 3)) 1+ 2 * * 2 3 ((1 + 2) * 3) (1 + 2) * 3 * ( (a / b) + ( (c - d) * e ) ) a / b + (c - d) * e + / a b * - c d e CIS210 95
96 Postfix and Prefix Notations Postfix and prefix notations do not need! Parentheses Operator precedence rules Operator association rules CIS210 96
97 Algebraic Expressions as Expression Trees Algebraic expressions involving binary operations can be represented by labeled binary trees. Expression trees represent algebraic expressions! Op Opd1 Op Opd2 Opd1 Opd2 CIS210 97
98 Algebraic Expression a / b + (c - d) * e ( (a / b) + ( (c - d) * e ) ) a b / c d - e * + + / a b * - c d e CIS210 98
99 Expression Tree for a / b + (c - d) * e + / * a b - e c d CIS210 99
100 Expression Tree ADT // exprtree.h // Class declarations for the linked implementation of the // Expression Tree ADT // class ExprTree; // Forward declaration of the ExprTree class class ExprTreeNode // Facilitator class for the ExprTree class { private: // Constructor ExprTreeNode ( char elem, ExprTreeNode *leftptr, ExprTreeNode *rightptr ); // Data members char dataitem; // Expression tree data item ExprTreeNode *left, // Pointer to the left child *right; // Pointer to the right child friend class ExprTree; }; CIS
101 Expression Tree ADT class ExprTree { public: // Constructor ExprTree (); // Destructor ~ExprTree (); // Expression tree manipulation operations void build (); // Build tree from prefix expression void expression () const; // Output expression in infix form float evaluate () const; // Evaluate expression void clear (); // Clear tree CIS
102 Expression Tree ADT // Output the tree structure -- used in testing/debugging void showstructure () const; ExprTree ( const ExprTree &valuetree ); // Copy constructor CIS
103 Expression Tree ADT private: // Recursive partners of the public member functions -- insert // prototypes of these functions here. void buildsub ( ExprTreeNode *&p ); void exprsub ( ExprTreeNode *p ) const; float evaluatesub ( ExprTreeNode *p ) const; void clearsub ( ExprTreeNode *p ); void showsub ( ExprTreeNode *p, int level ) const; // Data member ExprTreeNode *root; // Pointer to the root node }; CIS
104 Expression Tree ADT // // exprtree.cpp // ExprTreeNode:: ExprTreeNode ( char nodedataitem, ExprTreeNode *leftptr, ExprTreeNode *rightptr ) // Creates an expreesion tree node containing // data item nodedataitem, // left child pointer leftptr, and right child pointer rightptr. : dataitem(nodedataitem), left(leftptr), right(rightptr) {} CIS
105 Expression Tree ADT ExprTree:: ExprTree () // Creates an empty expression tree. : root(0) {} CIS
106 Expression Tree ADT // ExprTree:: ~ExprTree () // Frees the memory used by an expression tree. { clear(); } void ExprTree:: clear () // Removes all the nodes from an expression tree. { } clearsub(root); root = 0; CIS
107 Expression Tree ADT void ExprTree:: clearsub ( ExprTreeNode *p ) // Recursive partner of the clear() function. Clears the subtree // pointed to by p. { } if ( p!= 0 ) { clearsub(p->left); clearsub(p->right); delete p; } CIS
108 Expression Tree ADT void ExprTree:: build () // Reads a prefix expression (consisting of single-digit, nonnegative // integers and arithmetic operators) from the keyboard and // builds the corresponding expression tree. { buildsub(root); } CIS
109 Expression Tree ADT void ExprTree:: buildsub ( ExprTreeNode *&p ) // Recursive partner of the build() function. Builds a subtree and // sets p to point to its root. { char ch; // Input operator or number cin >> ch; p = new ExprTreeNode(ch,0,0); // Link in node if (!isdigit(ch) ) // Operator -- construct subtrees } { } buildsub(p->left); buildsub(p->right); CIS
110 Expression Tree ADT void ExprTree:: expression () const // Outputs the corresponding arithmetic expression in fully // parenthesized infix form. { exprsub(root); } CIS
111 Expression Tree ADT void ExprTree:: exprsub ( ExprTreeNode *p ) const // Recursive partner of the expression() function. // Outputs the subtree pointed to by p. { if ( p!= 0 ) { if (!isdigit(p->dataitem) ) cout << '('; exprsub(p->left); cout << p->dataitem; exprsub(p->right); if (!isdigit(p->dataitem) ) cout << ')'; } } CIS
112 Expression Tree ADT float ExprTree:: evaluate () // Returns the value of the corresponding arithmetic expression. { // Requires that the tree is not empty return evaluatesub(root); } CIS
113 Expression Tree ADT float ExprTree:: evaluatesub ( ExprTreeNode *p ) const // Recursive partner of the evaluate() function. Returns the value of // subtree pointed to by p. { float l, r, // Intermediate results result; // Result returned if ( isdigit(p->dataitem) ) result = p->dataitem - '0'; // Convert from char to number else { l = evaluatesub(p->left); // Evaluate subtrees r = evaluatesub(p->right); switch ( p->dataitem ) // Combine results { case '+' : result = l + r; break; case '-' : result = l - r; break; case '*' : result = l * r; break; case '/' : result = l / r; } } return result; CIS }
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