Prefix to Infix Conversion

Infix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2).
Example : (A+B) * (C-D)

Prefix : An expression is called the prefix expression if the operator appears in the expression before the operands. Simply of the form (operator operand1 operand2).
Example : *+AB-CD (Infix : (A+B) * (C-D) )

Given a Prefix expression, convert it into a Infix expression.
Computers usually does the computation in either prefix or postfix (usually postfix). But for humans, its easier to understand an Infix expression rather than a prefix. Hence conversion is need for human understanding.


Input :  Prefix :  *+AB-CD
Output : Infix : ((A+B)*(C-D))

Input :  Prefix :  *-A/BC-/AKL
Output : Infix : ((A-(B/C))*((A/K)-L)) 

Algorithm for Prefix to Infix:

  • Read the Prefix expression in reverse order (from right to left)
  • If the symbol is an operand, then push it onto the Stack
  • If the symbol is an operator, then pop two operands from the Stack
    Create a string by concatenating the two operands and the operator between them.
    string = (operand1 + operator + operand2)
    And push the resultant string back to Stack
  • Repeat the above steps until end of Prefix expression.




// CPP Program to convert prefix to Infix
#include <iostream>
#include <stack>
using namespace std;
// function to check if character is operator or not
bool isOperator(char x) {
  switch (x) {
  case '+':
  case '-':
  case '/':
  case '*':
    return true;
  return false;
// Convert prefix to Infix expression
string preToInfix(string pre_exp) {
  stack<string> s;
  // length of expression
  int length = pre_exp.size();
  // reading from right to left
  for (int i = length - 1; i >= 0; i--) {
    // check if symbol is operator
    if (isOperator(pre_exp[i])) {
      // pop two operands from stack
      string op1 =;   s.pop();
      string op2 =;   s.pop();
      // concat the operands and operator
      string temp = "(" + op1 + pre_exp[i] + op2 + ")";
      // Push string temp back to stack
    // if symbol is an operand
    else {
      // push the operand to the stack
      s.push(string(1, pre_exp[i]));
  // Stack now contains the Infix expression
// Driver Code
int main() {
  string pre_exp = "*-A/BC-/AKL";
  cout << "Infix : " << preToInfix(pre_exp);
  return 0;



Infix : ((A-(B/C))*((A/K)-L))

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.