/// <summary>
/// Calculates the value of the binary tree
/// </summary>
/// <param name="curNode">Either the root of the equation or the branch</param>
/// <returns>double.Nan if the expression is invalid. Otherwise returns the value of the expression</returns>
public double CalculateTree(Node curNode)
{
//For the binary tree to be properly constructed, all of the prefix characters should be used
//If the count counter does not equal to the lenght of our prefix string, it proves that the binary tree was not properly built
if (count != prefix.Count())
{
return(double.NaN);
}
//Catches any errors within the equation such as;
//Dividing by zero
//Converting non numeric values; Example: "a", " ", "$"
try
{
//Executes all calculations here
//Calls CalculateTree until it returns the default scenario
//Default scenario converts the string into a double, operators would have been already be dealt with
//Anything other than a numberic value would mean the binary tree was not construction properly
switch (curNode.GetCargo())
{
case ("+"):
answer = +CalculateTree(curNode.GetNextL()) + CalculateTree(curNode.GetNextR());
return(answer);
case ("-"):
answer = +CalculateTree(curNode.GetNextL()) - CalculateTree(curNode.GetNextR());
return(answer);
case ("/"):
answer = +CalculateTree(curNode.GetNextL()) / CalculateTree(curNode.GetNextR());
return(answer);
case ("*"):
answer = +CalculateTree(curNode.GetNextL()) * CalculateTree(curNode.GetNextR());
return(answer);
case ("^"):
answer = +Math.Pow(CalculateTree(curNode.GetNextL()), CalculateTree(curNode.GetNextR()));
return(answer);
default:
return(Convert.ToDouble(curNode.GetCargo()));
}
}
catch (Exception exp)
{
return(double.NaN);
}
}
/// <summary>
/// Sets the left and right nodes of our curNode to the first two letters of our prefix accordingly
/// Uses recursives methods if certain conditions are met
/// </summary>
/// <param name="curNode">Either the root of the equation or a branch</param>
/// <param name="Prefix">Prefix input from the text file</param>
public void TreeBuilder(Node curNode, string Prefix)
{
//Create a new node for the current nodes left branch to
Node leftNode = new Node(Prefix.Substring(count, 1));
curNode.SetLeft(leftNode);
count++;
//Call TreeBuilder again if the left branch is an operation
//Sets the TreeBuilder parameter to the current node's left branch asa well as the same prefix
if (leftNode.GetCargo() == "+" || leftNode.GetCargo() == "-" || leftNode.GetCargo() == "*" || leftNode.GetCargo() == "/" || leftNode.GetCargo() == "^")
{
TreeBuilder(curNode.GetNextL(), Prefix);
}
//Create a new node for the current nodes right branch to
Node rightNode = new Node(Prefix.Substring(count, 1));
curNode.SetRight(rightNode);
count++;
//Call TreeBuilder again if the right branch is an operation
//Sets the TreeBuilder parameter to the current node's right branch asa well as the same prefix
if (rightNode.GetCargo() == "+" || rightNode.GetCargo() == "-" || rightNode.GetCargo() == "*" || rightNode.GetCargo() == "/" || rightNode.GetCargo() == "^")
{
TreeBuilder(curNode.GetNextR(), Prefix);
}
}
/// <summary>
/// Converts the binary tree of the prefix equation into an infix equation
/// </summary>
/// <param name="curNode">Either the root of the equation or the branch</param>
/// <returns>The infix equation of the binary tree</returns>
public string InfixDisplay(Node curNode)
{
//Calls InfixDisplay on both left and right branches if the current node is an operation
//Otherwise, return the current nodes cargo with the appropriate spacing
if (curNode.GetCargo() == "+" || curNode.GetCargo() == "-" || curNode.GetCargo() == "*" || curNode.GetCargo() == "/" || curNode.GetCargo() == "^")
{
return("(" + InfixDisplay(curNode.GetNextL()) + " " + curNode.GetCargo() + " " + InfixDisplay(curNode.GetNextR()) + ")");
}
else
{
return(" " + curNode.GetCargo() + " ");
}
}