Infix to Postfix Conversion using Stack in C++ (original) (raw)

Last Updated : 23 Jul, 2025

Infix expression is a common way of writing mathematical expressions where operators are written between the operands whereas postfix is a type of expression in which a pair of operands is followed by an operator. In this article, we will learn how to use a stack to convert an infix expression to a postfix expression in C++.

**Example:

*Input:
Infix expression: “A+BC”

*Output:
Postfix expression: “ABC+”

Infix to Postfix Conversion using Stack in C++

To convert an infix expression to a postfix expression using a std::stack in C++, we can follow the below approach:

**Approach:

• Create an empty stack for storing operators and a string for storing the result.
• Scan the infix expression from left to right.
• If the scanned character is an operand, append it to the result.
• If the scanned character is an operator, pop operators from the stack to the result until the top of the stack has an operator of lower precedence or the stack is empty, then push the scanned operator onto the stack.
• If the scanned character is ‘(’, push it onto the stack.
• If the scanned character is ‘)’, pop operators from the stack to the result until ‘(’ is encountered, and pop ‘(’ from the stack.
• After all characters are scanned, pop the remaining operators from the stack to the result.

C++ Program to Convert an Infix Expression to a Postfix Expression using a Stack

The following program illustrates how we can convert an infix expression to a postfix expression using a stack in C++.

C++ `

// C++ Program to illustrate how to use a stack to convert // an infix expression to a postfix expression #include #include #include using namespace std;

// Function to check the precedence of operators int precedence(char op) { if (op == '+' || op == '-') return 1; if (op == '*' || op == '/') return 2; return 0; }

// Function to convert infix expression to postfix // expression string infixToPostfix(string infix) { stack st; string postfix = ""; for (int i = 0; i < infix.length(); i++) { char c = infix[i];

    // If the scanned character is an operand, add it to
    // output string.
    if (isalnum(c))
        postfix += c;

    // If the scanned character is an '(', push it to
    // the stack.
    else if (c == '(')
        st.push('(');

    // If the scanned character is an ')', pop and to
    // output string from the stack until an '(' is
    // encountered.
    else if (c == ')') {
        while (st.top() != '(') {
            postfix += st.top();
            st.pop();
        }
        st.pop();
    }

    // If an operator is scanned
    else {
        while (!st.empty()
               && precedence(c)
                      <= precedence(st.top())) {
            postfix += st.top();
            st.pop();
        }
        st.push(c);
    }
}

// Pop all the remaining elements from the stack
while (!st.empty()) {
    postfix += st.top();
    st.pop();
}

return postfix;

}

int main() { string infix = "A+B*C"; cout << "Infix Expression: " << infix << endl; cout << "Postfix Expression: " << infixToPostfix(infix) << endl; return 0; }

`

Output

Infix Expression: A+BC Postfix Expression: ABC+

**Time Complexity: O(N), here N is the length of infix expression.
**Auxiliary Space: O(N)