private Node <OperateObj> buildParseTreeHelp()//保证调用前栈顶一定是运算符
{
if (parseStack2.Count == 0)
{
return(null);
}
if (parseStack2.Peek().objType != 1)
{
return(new Node <OperateObj>(parseStack2.Pop()));
}
OperateObj parentValue = parseStack2.Pop();
Node <OperateObj> parentNode = null;
Node <OperateObj> leftNode = null;
Node <OperateObj> rightNode = null;//其实是指向一个parent的指针
if (parseStack2.Count == 0)
{
currErrorType = 4;
return(null);
}
switch (parseStack2.Peek().objType) //为右孩子操心
{
case 1: //操作符
rightNode = buildParseTreeHelp();
break;
case 2: //操作数
rightNode = new Node <OperateObj>(parseStack2.Pop()); //左子的左右子为null
break;
default:
break;
}
if (parseStack2.Count == 0)
{
currErrorType = 4;
return(null);
}
switch (parseStack2.Peek().objType) //为左孩子操心
{
case 1: //操作符
leftNode = buildParseTreeHelp();
break;
case 2: //操作数
leftNode = new Node <OperateObj>(parseStack2.Pop());
break;
default:
break;
}
parentNode = new Node <OperateObj>(parentValue, leftNode, rightNode);
return(parentNode);
}
//构造解析树 迭代！ 递归自下至上递归？ 没见过 大概不行
//迭代要遍历+操作符栈+操作数栈 递归只管当下
//建树最好从顶向下建
private bool buildParseTree() //先将RPN整到一个解析栈中 再通过栈递归建树
{
if (l.Count == 0)
{
return(false);
}
if (currErrorType != 0)
{
return(false);
}
parseStack.Clear();
parseStack2.Clear();
// parseTree.clear();
int index = 0;
while (index != l.Count)//如果表达式格式非法 (应该在界面层防止这个问题发生） 建栈会如何？
{
OperateObj obj = l.ElementAt <OperateObj>(index);
switch (obj.objType)
{
case 1: //操作符
Operator optTemp = (Operator)obj;
//Operator optStackTop;
//if (parseStack.Count==0)
// optStackTop=null;
//else
// optStackTop=(Operator)parseStack.Peek();
if (parseStack.Count == 0)
{
parseStack.Push(obj);
}
else
{
while (parseStack.Count != 0 && optTemp.getPriority() <= parseStack.Peek().getPriority())
{
parseStack2.Push(parseStack.Pop());
}
parseStack.Push(obj);
}
break;
case 2: //操作数
parseStack2.Push(obj);
break;
case 3: //括号
Bracket objTemp = (Bracket)obj;
if (objTemp.isLeft) //左括号压栈
{
parseStack.Push(obj);
}
else //右括号一直弹栈直到左括号出栈
{
while (parseStack.Peek().objType != 3) //不停弹栈
{
parseStack2.Push(parseStack.Pop());
// Operator objTemp2 = (Operator)parseStack.Pop();
//leftNode=new Node<OperateObj>()
}
parseStack.Pop();
}
break;
default:
break;
}
index++;
//return new BinaryTree<OperateObj>(ln); //这一行似乎没有必要
//parse
}
while (parseStack.Count != 0)
{
parseStack2.Push(parseStack.Pop());
}
parseTree = new BinaryTree <OperateObj> (buildParseTreeHelp());
if (CurrentErrorType == 4)
{
return(false);
}
return(true);
}
