cpp parse/explode expression into expression without brackets

Question

I'm trying to build a "parser" in C++, that expands/explodes a formula into the equivalent without brackets, but im super stuck. A hint in the right direction would be super cool!

Example input:

1: "a and (b or c)"
2: "a and (b or (c and d))"

Desired output:

1: "a and b or a and c"
2: "a and b or a and c and d"

Evaluation the formulas is already done. But transforming the formula is getting me really frustrated. How would you solve this problem?

Thank you in advance!

Update: Thank you for the reply HolyBlackCat!

While searching for a solution I encountered the shunting-yard algorithm aswell. The following code is an implementation of that, and returns the infix notation correctly.

/* C++ implementation to convert
infix expression to postfix*/

#include <bits/stdc++.h>
using namespace std;

// Function to return precedence of operators
int prec(char c)
{
    if (c == '^')
        return 3;
    else if (c == '&' || c == '*')
        return 2;
    else if (c == '+' || c == '-')
        return 1;
    else
        return -1;
}

// Convert infix expression to postfix expression
void infixToPostfix(string s)
{

    stack<char> st; // C++ built in stack
    string result;

    for (int i = 0; i < s.length(); i++)
    {
        char c = s[i];

        // If the scanned character is
        // an operand, add it to output string.
        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))
            result += 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() != '(')
            {
                result += st.top();
                st.pop();
            }
            st.pop();
        }

        // If an operator is scanned
        else
        {
            while (!st.empty() && prec(s[i]) <= prec(st.top()))
            {
                if (c == '^' && st.top() == '^')
                    break;
                else
                {
                    result += st.top();
                    st.pop();
                }
            }
            st.push(c);
        }
    }

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

    cout << result << endl;
}

int main()
{
    string exp = "a*(c+b);";
    infixToPostfix(exp);
    return 0;
}

Outputs: acb+*

But still, I foresee how to turn it back into postfix notation with "exploded" parenthesis. Any push in the right direction would be awesome!