# Postfix to Infix Conversion

Postfix to Infix conversion” is a classic example of  stack data structure. Stacks can be used to convert given postfix expression to corresponding infix expression.

Operator: Operator are symbols that instruct the computer to perform simple and single tasks. Examples of operators includes + (Addition), – (Subtraction), * (multiplication),… and many more.

Operand: Operands are values or variable on which operator works its tasks. Examples of Operand includes “a”, “b”, 23, 12,.. and many more.

Steps to convert Postfix to Infix Expression

1. Scan the postfix expression from left to right.
2. Initialize an empty string stack.
3. If the scanned character is operand, push it into stack.
4. Else if the scanned character is operator, pop two strings from stack, namely, temp1 and temp2, and  push: (temp2 operator temp1) into stack.
5. Repeat steps from 3 to 4 until all the characters from the string are scanned.
6. In the end, only one valid infix string will be present in the stack, pop it and return it.

Example

Suppose, we want to convert the following postfix expression to infix expression

Algorithm for converting postfix expression to infix expression

```Algo postfix_2_infix (postfix)
{    // input- valid postfix expression
//output- equivalent infix expression
1.    Createstack (stack).
2.    i=0
3.    loop(i<sizeof(stack))
{
a.if postfix[i] is an operator
{
operand2=popstack().
operand1=popstack().
temp= concatenate “(“+ operand1 + postfix[i] + operand2 + “)”.
pushstack (temp).
}
b.else if postfix[i] is an operand
Then push postfix[i] into stack.
c.Increment  i  with 1.
}
4.    infix= popstack().
5.    Return infix.
}// END OF ALGO.

```

C++ Program for converting postfix expression to infix expression

```#include <iostream>
#include<string>
#define sizes 100
using namespace std;
class stack
{
string item;
int top;
public:
stack()
{
top=-1;
}
void push(string str)
{
if(top==sizes-1)
{
cout<<"stack overflow!!\n";
return;
}
top++;
item[top]=str;
}
string pop()
{
int i;
string temp;
if(top==-1)
{
cout<<"stack underflow!!\n";
return "abc";
}
temp = item[top];
top--;
return temp;
}
};
int main(int argc, char** argv)
{
int i,j=0;
stack st;
string postfix,infix;
cout<<"Enter postfix expression:\n";
cin>>postfix;
for(i=0;i<postfix.size();i++)
{
if(postfix[i]=='+' || postfix[i]=='-' || postfix[i]=='*' || postfix[i]=='/' || postfix[i]=='^')
{
string temp,op1,op2;
op2=st.pop();
op1=st.pop();
temp='('+op1+postfix[i]+op2+')';
st.push(temp);
}
else
{
string flag;
flag=flag+postfix[i];
st.push(flag);
}
}
cout<<"The equivalent infix expression is:\n"<<st.pop();
return 0;
}

```

```OUTPUT:
Enter postfix expression:
abc*de-/+
The equivalent infix expression:
((a+((b*c)/(d-e))))```

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