1 CS 163 Data Structures Chapter 9 Building, Printing Binary Trees Herbert G. Mayer, PSU Status 5/21/2015

2 Syllabus Arithmetic Expressions and Trees Arithmetic Expressions and Trees Infix Without Parentheses Infix Without Parentheses Infix With Parentheses Infix With Parentheses Postfix Without Parentheses Postfix Without Parentheses Prefix Without Parentheses Prefix Without Parentheses Interesting Examples Interesting Examples Use of Postfix Use of Postfix Building Trees Building Trees Tree of Ints Tree of Ints

3 Arithmetic Expressions and Trees Three typical notations for dyadic operations: Three typical notations for dyadic operations: Infix notation: write as the first the left operand, reading left-to- right, then list the dyadic operator, finally list the right operand Infix notation: write as the first the left operand, reading left-to- right, then list the dyadic operator, finally list the right operand For CPU: Order will not work for code emission, as the CPU needs both operands for processing the operator For humans: requires parentheses for proper operator precedence Note exception: programming language APL Postfix notation: write left operand first, then list the right operand, finally the operator Postfix notation: write left operand first, then list the right operand, finally the operator This order will work for code emission, as operator has both operands available at processing time Needs no parentheses, and still obeys operator precedence Postfix notation AKA Polish Postfix, after Jan Łukasiewicz, 1920 Prefix notation: First list the operator, next the first (left) operand, finally the second (right) operand Prefix notation: First list the operator, next the first (left) operand, finally the second (right) operand

4 Arithmetic Expressions and Trees a + x ^ c + a^ xc Infix:( a + ( x ^ c ) ) Postfix:a x c ^ + Prefix:+ a ^ x c ^ stands for exponentiation, with highest precedence: higher than * or /

5 Arithmetic Expressions and Trees ( x – a ) / b / - b ax Infix:( ( x – a ) ) / b Postfix:x a – b / Prefix:/ – x a b / stands for division operator, with higher precedence than, say, –

6 Arithmetic Expressions and Trees a ^ ( b – c ) / d / ^ d -a Infix:( ( a ^ ( b – c ) ) / d ) Postfix:a b c - ^ d / Prefix:/ ^ a – b c d bc

7 Data Structure to Print Trees // node has class: literal, identifier, or operator. // Parenthesized expressions have been reduced: no ( ) typedef enum { Literal, Identifier, Operator } NodeClass; typedef struct NodeType * NodePtr; // forward // actual node structure; using the forward pointers typedef struct NodeType { NodeClass Class; // 3 classes. Not C++ ‘class’ NodeClass Class; // 3 classes. Not C++ ‘class’ char Symbol; // stores ident or small literal char Symbol; // stores ident or small literal int LitVal; // if Class == Literal: its value int LitVal; // if Class == Literal: its value NodePtr Left; // left subtree NodePtr Left; // left subtree NodePtr Right; // right subtree NodePtr Right; // right subtree } s_node_tp;

8 Infix Without Parentheses // Print in infix notation without parentheses ( ) void Print_No_Paren( NodePtr Root ) { // Print_No_Paren if ( Root ) { Print_No_Paren ( Root->Left ); if ( Root->Class == Literal ) { printf( "%d", Root->LitVal ); }else{ printf( "%c", Root->Symbol ); } //end if Print_No_Paren ( Root->Right ); } //end if } //end Print_No_Paren Input: ( a + x ) / b prints as: a + x / b  misleading

9 Infix With Parentheses // Print in infix notation with parentheses ( and ) void Print_Infix( NodePtr Root ) { // Print_Infix if ( Root ) { if ( Root->Class == Operator ) { printf( "(" ); } //end if Print_Infix( Root->Left ); if ( Root->Class == Literal ) { printf( "%d", Root->LitVal ); }else{ printf( "%c", Root->Symbol ); } //end if Print_Infix( Root->Right ); if ( Root->Class == Operator ) { printf( ")" ); } //end if } //end Print_Infix Input: ( a + x ) / b prints as: ( ( a + x ) / b ) -- OK

10 Postfix Without Parentheses // Print in Polish Postfix notation, no parentheses void Print_Postfix( NodePtr Root ) { // Print_Postfix if ( Root ) { Print_Postfix( Root->Left ); Print_Postfix( Root->Right ); if ( Root->Class == Literal ) { printf( "%d", Root->LitVal ); }else{ printf( "%c", Root->Symbol ); } //end if } //end Print_Postfix Input: a ^ ( b – c ) / d prints as: a b c - ^ d / -- OK

11 Prefix Without Parentheses // Prefix: operator executes when 2 operands found void Print_Prefix( NodePtr Root ) { // Print_Prefix if ( Root ) { if ( Root->Class == Literal ) { printf( "%d", Root->LitVal ); }else{ printf( "%c", Root->Symbol ); } //end if Print_Prefix ( Root->Left ); Print_Prefix ( Root->Right ); } //end if } //end Print_Prefix Input: ( a + x ) / b prints as: / + a x b -- OK

12 Interesting Examples Input 1:a + b * c ^ ( x – 2 * d ) / ( e – f ) Infix:( a + ( ( b * ( c ^ ( x – ( 2 * d ) ) ) ) / ( e – f ) ) ) Postfix:a b c x 2 d * - ^ * e f - / + Prefix:+ a / * b ^ c – x * 2 d – e f Input 2:4 / x ^ ( k – l / m ) * 8 * x - & 9 + n Infix:( ( ( ( ( 4 / ( x ^ ( k - ( l / m ) ) ) ) * 8 ) * x ) - ( & 9 ) ) + n ) Postfix:4 x k l m / - ^ / 8 * x * 9 & - n + Prefix:+ - * * / 4 ^ x – k / l m 8 x & 9 n

13 Use of Postfix Postfix, AKA Polish Postfix notation is a natural for code generation, targeted for stack machines Postfix, AKA Polish Postfix notation is a natural for code generation, targeted for stack machines Operands are needed first: Two for dyadic, or one for monadic operations Operands are needed first: Two for dyadic, or one for monadic operations Once generated and available on stack, stack machine can execute the next operation Once generated and available on stack, stack machine can execute the next operation Easy for compiler writer, natural for stack machine Easy for compiler writer, natural for stack machine Stack poor for execution, as all references are through memory: top of stack Stack poor for execution, as all references are through memory: top of stack Even a GPR architecture needs both operands available somewhere (in regs) to execute operator Even a GPR architecture needs both operands available somewhere (in regs) to execute operator

14 Building Trees Now you understand: Trees constitute inherently recursive data structures Now you understand: Trees constitute inherently recursive data structures You learned how to traverse them, and in various orders You learned how to traverse them, and in various orders Now we learn how to build them Now we learn how to build them Initially trees may be unbalanced Initially trees may be unbalanced Worst case, they may be so-called left- spine (or right-spine) trees, i.e. no different from linear lists, but more costly to process Worst case, they may be so-called left- spine (or right-spine) trees, i.e. no different from linear lists, but more costly to process

15 Building Trees Let tree nodes be simple int data structures Let tree nodes be simple int data structures Each node holds an integer value named info Each node holds an integer value named info Info numbers don’t need to be unique Info numbers don’t need to be unique Provide data from input files, or hard-coded Provide data from input files, or hard-coded Since node numbers may be repeated, each node includes a count Since node numbers may be repeated, each node includes a count Let the left and right subtrees be named smaller and greater, alluding to info ordering Let the left and right subtrees be named smaller and greater, alluding to info ordering Thus we have sufficient information to define the node data structure in C++ Thus we have sufficient information to define the node data structure in C++

16 Building Trees, Node Type typedef struct node_tp * node_ptr_tp; typedef struct node_tp { int info;// the node! int info;// the node! int count;// how many? int count;// how many? node_ptr_tp smaller;// left subtree node_ptr_tp smaller;// left subtree node_ptr_tp greater; // and right node_ptr_tp greater; // and right } str_node_tp; node_ptr_tp root = NULL; // key Datum

17 Allocate Node Off Heap //use C malloc() or C++ new operator node_ptr_tp make_node( int info ) { // make_node node_ptr_tp node= new str_node_tp; if( node ) { node->info= info; node->info= info; node->smaller= NULL; node->smaller= NULL; node->greater= NULL; node->greater= NULL; node->count= 1; node->count= 1;}else{ error( ”...” ); } //end if return node; return node; } //end make_node

18 Tree of Ints Tree is pointed to by root, initially nil Tree is pointed to by root, initially nil Type of root is pointer to node, in C++ node_ptr_tp Type of root is pointer to node, in C++ node_ptr_tp When the next info element is searched in an empty tree, i.e. if root == NULL, new node is inserted, count set to 1, root points to that new node, and root is returned When the next info element is searched in an empty tree, i.e. if root == NULL, new node is inserted, count set to 1, root points to that new node, and root is returned Done by node_ptr_tp function insert() Done by node_ptr_tp function insert() Else, since tree is not empty, the current info is compared; if yields a match, count is incremented and root to that existing node is returned Else, since tree is not empty, the current info is compared; if yields a match, count is incremented and root to that existing node is returned

19 Tree of Ints Else, if there is no match, traverse the left or the right subtree, depending on whether root->greater or root->smaller Else, if there is no match, traverse the left or the right subtree, depending on whether root->greater or root->smaller I.e., whether info > root->info I.e., whether info > root->info Or whether info info Or whether info info And recurse! And recurse!

20 Tree of Ints // Given “info”, insert in tree: “root” node_ptr_tp insert( node_ptr_tp & root, int info ) { // insert if ( NULL == root ) {// why in this order? root = make_node( info ); }else if( info > root->info ) { }else if( info > root->info ) { root->greater = insert( root->greater, info ); root->greater = insert( root->greater, info ); }else if( info info ) { }else if( info info ) { root->smaller = insert( root->smaller, info ); root->smaller = insert( root->smaller, info ); }else{ }else{ root->count++; root->count++; } //end if } //end if return root; return root; } //end insert

21 Traverse Tree of Ints In-order When the complete tree is constructed, it is pointed to by global root, of node_ptr_tp When the complete tree is constructed, it is pointed to by global root, of node_ptr_tp To traverse, we can use pre-order, post- order or in-order To traverse, we can use pre-order, post- order or in-order In-order handles left subtree first In-order handles left subtree first Then looks at node Then looks at node And finally traverses right subtree And finally traverses right subtree

22 Traverse Tree of Ints In-order // traverse binary tree, root, in in-order void in_order( node_ptr_tp root ) { // in_order if ( root ) { if ( root ) { in_order( root->smaller ); in_order( root->smaller ); printf( "%d(%d) ", root->info, printf( "%d(%d) ", root->info, root->count ); in_order( root->greater ); in_order( root->greater ); } //end if } //end if } //end in_order

23 Sample Tree Traversal Let’s build a binary tree of ints, pointed to by root Let’s build a binary tree of ints, pointed to by root The input sequence we consider is given here: The input sequence we consider is given here: We build the tree as shown above We build the tree as shown above Tracking the count of each integer value, by giving this in parentheses Tracking the count of each integer value, by giving this in parentheses E.g. -11(3) would mean: integer value -11 is in the tree and did occur 3 times in the input provided E.g. -11(3) would mean: integer value -11 is in the tree and did occur 3 times in the input provided

24 Sample Tree Traversal Hard-coded input of data to build tree: Hard-coded input of data to build tree: // enter some arbitrary ints void get_data( void ) { // get_data for ( int node = -5; node < 20; node += 3 ) { for ( int node = -5; node < 20; node += 3 ) { root = insert( root, node ); root = insert( root, node ); root = insert( root, node - 9 ); root = insert( root, node - 9 ); // printf( "%d %d ", node, node - 9 ); // printf( "%d %d ", node, node - 9 ); } //end for } //end for } //end get_data

25 Sample Tree Traversal int main() { // main printf( "Entering fixed list of: " ); get_data( ); get_data( ); printf( "\n in order : " ); printf( "\n in order : " ); in_order( root ); in_order( root ); printf( "\n" ); printf( "\n" ); return 0; return 0; } //end main

26 Sample Output in order : -14(1) -11(1) -8(1) -5(2) -2(2) 1(2) 4(2) 7(2) 10(2) 13(1) 16(1) 19(1)

27 References Łukasiewicz: Łukasiewicz: