postifx to infix binary expression tree (parenthesis) using stack

Question

I have a program (c++) I'm writing where I'm converting a postfix expression to infix

Example Postfix: ab + cde + * * Converted to infix: ( a + b ) * ( c * ( d + e ) )

You do this by traversing "inorderly" through the binary expression tree

I have a recursiving function that prints out the expression properly, but I can't quite figure out how to place the paranthesis in the correct place. My best attempt was the result ( a + b ) * ( c * ( d + e //but I can't place the end parenthesis correctly without messing up other parts

This is the function that produced this result (and I'm pretty sure I need to rethink my strategy for consistent results:

//isHigher is a lamba function that checks for higher precedence operators (*, /)
//isoperator checks if its an operator (+, - , / * )
void BET::printInfixExpression(BinaryNode *n)
{
  if(n->left != NULL)
  {
    if(isOperator(n->left->element) && isHigher(n->element) && !isHigher(n->left->element))
      cout << "( ";
    if(isHigher(n->element) && !isOperator(n->left->element))
      cout << "( ";
    printInfixExpression(n->left);
  }
  if(isHigher(n->element) && isHigher(n->right->element))
    cout << ") ";
  cout << n->element << " ";
  if(isHigher(n->element) && isOperator(n->right->element) && !isHigher(n->right->element))
    cout << "( ";
  if(n->right != NULL)
  {
    printInfixExpression(n->right);
  }
}

This is the original function that outputs an infix with no parenthesis:

void BET::printInfixExpression(BinaryNode *n)
{
  if(n->left != NULL)
  {
    printInfixExpression(n->left);
  }
  cout << n->element << " ";

  if(n->right != NULL)
  {
    printInfixExpression(n->right);
  }
}

So my problem is getting the parenthesis placed correctly. Any help would be greatly appreciated, It's the weekend so my TA's/teacher haven't gotten back to me.

edit: Redundant parenthesis is not allowed. It has to be appropriately placed when necessary but not when it isn't necessary.

Answer 1

You don't need to know the operator precedence. It is already implicit in the postfix. You just need to put parentheses around every operand-operator-operand tuple as you output them. So for example, ab+ becomes (a+b), abc++ becomes (a+(b+c)). So just output a "(" at the beginning of printInfixExpression() and a ")" at the end of it. You will get redundant parentheses, but I don't see anything in your question about suppressing those.