Name:: Muhammad Usman

Roll no:: 6519

Class:: Mcs eve 2

Data structure Assignment

Q:Evaluate postfix expression?

#include <iostream>

#include <stack>

using namespace std;

// Function to evaluate given postfix expression

int postfixEval(string exp)

// create an empty stack

stack<int> stack;

// traverse the given expression

for (int i = 0; i < exp.length(); i++)

// if current char is an operand, push it to the stack

if (exp[i] >= '0' && exp[i] <= '9') {

stack.push(exp[i] - '0');

// if current char is an operator

else

// pop top two elements from the stack

int x = stack.top();

stack.pop();
// if (stack.empty())

int y = stack.top();

stack.pop();

// evaluate the expression x op y, and push the

// result back to the stack

if(exp[i] == '+')

stack.push(y + x);

else if(exp[i] == '-')

stack.push(y - x);

else if(exp[i] == '*')

stack.push(y * x);

else if(exp[i] == '/')

stack.push(y / x);

// At this point, the stack is left with only one element

// i.e. expression result

return stack.top();

int main()

string exp;

cout<<"enter any postfix expreesion "<<endl;

cin>>exp;

cout << postfixEval(exp);

return 0;

Q:Convert Infix to postfix?

#include<iostream>
#include<stack>

using namespace std;

void Postfix(char *a)

{ stack <char> s;

char output[50],t;

for(int i=0;a[i]!='\0';i++)

{ char ch = a[i];

switch(ch)

case '^':

case '-':

case '+':

case '/':

case '*': s.push(ch);

break;

case ')': t=s.top();

s.pop();

cout<<t;

break;

if (isalpha(ch))

cout<<ch;

int main()

char a[15];

cout<<"enter any infix expression"<<endl;

for(int i=0;i<15;i++)
 {

cin>>a[i];

Postfix(a);

return 0;

Q:Convert Infix to prefix?

#include <iostream>

#include<stack>

using namespace std;

// Function to check if given character is

// an operator or not.

bool isOperator(char c)

return (!isalpha(c) && !isdigit(c));

// Function to find priority of given

// operator.

int getPriority(char C)

if (C == '-' || C == '+')

return 1;

else if (C == '*' || C == '/')

return 2;

else if (C == '^')

return 3;
 return 0;

// Function that converts infix

// expression to prefix expression.

string infixToPrefix(string infix)

// stack for operators.

stack<char> operators;

// stack for operands.

stack<string> operands;

for (int i = 0; i < infix.length(); i++) {

// If current character is an

// opening bracket, then

// push into the operators stack.

if (infix[i] == '(') {

operators.push(infix[i]);

// If current character is a

// closing bracket, then pop from

// both stacks and push result

// in operands stack until

// matching opening bracket is

// not found.

else if (infix[i] == ')') {

operators.top() != '(') {

// operand 1

string op1 = operands.top();

operands.pop();

// operand 2

string op2 = operands.top();

operands.pop();

// operator

char op = operators.top();

operators.pop();

// Add operands and operator

// in form operator +

// operand1 + operand2.

string tmp = op + op2 + op1;

operands.push(tmp);

// Pop opening bracket from

// stack.

operators.pop();

// If current character is an

// operand then push it into

else if (!isOperator(infix[i])) {

operands.push(string(1, infix[i]));

// If current character is an

// operator, then push it into

// operators stack after popping

// high priority operators from

// operators stack and pushing

// result in operands stack.

else {

while (!operators.empty() &&

getPriority(infix[i]) <=

getPriority(operators.top())) {

string op1 = operands.top();

operands.pop();

string op2 = operands.top();

operands.pop();

char op = operators.top();

operators.pop();

string tmp = op + op2 + op1;

operands.push(tmp);

}
 operators.push(infix[i]);

// Pop operators from operators stack

// until it is empty and add result

// of each pop operation in

// operands stack.

while (!operators.empty()) {

string op1 = operands.top();

operands.pop();

string op2 = operands.top();

operands.pop();

char op = operators.top();

operators.pop();

string tmp = op + op2 + op1;

operands.push(tmp);

// Final prefix expression is

// present in operands stack.

return operands.top();

// Driver code

int main()
{

string s = "(A-B/C)*(A/K-L)";

cout << infixToPrefix(s);

return 0;

}}

Q:convert postfix to Infix?

#include <bits/stdc++.h>

using namespace std;

bool isOperand(char x)

   return (x >= 'a' && x <= 'z') ||

          (x >= 'A' && x <= 'Z');

// Get Infix for a given postfix

// expression

string getInfix(string exp)

    stack<string> s;

    for (int i=0; exp[i]!='\0'; i++)

    {

        // Push operands

        if (isOperand(exp[i]))

        {

           string op(1, exp[i]);

           s.push(op);

        }
  

        // We assume that input is

        // a valid postfix and expect

        // an operator.

        else

        {

            string op1 = s.top();

            s.pop();

            string op2 = s.top();

            s.pop();

            s.push("(" + op2 + exp[i] +

                   op1 + ")");

        }

    }

    // There must be a single element

    // in stack now which is the required

    // infix.

    return s.top();

// Driver code

int main()

    string exp = "ab*c+";

    cout << getInfix(exp);

    return 0;