有足够的资源如何转换前pression树为后缀符号,它并不难。
There are enough resources on how to convert an expression tree into postfix notation, and it's not that hard.
不过,我要解析后缀EX pression到前pression树。
But I have to parse a postfix expression into an expression tree.
这位前pression是:
The expression is:
一个2 ^ 2 A * B * - B 2 ^ + A B - /
A 2 ^ 2 A * B * - B 2 ^ + A B - /
我真的没有一个线索如何跨preT前pression。是否有人对如何PROCES这个线索?
I don't really have a clue on how to interpret the expression. Does someone has a clue on how to proces this?
创建一个包含节点,可能是树的一部分一叠
Create a stack containing nodes that could be part of a tree
在堆栈(A,2,B等是操作数)为叶节点推送操作数,没有绑定到任何树在任何方向 对于运营商,流行的必要的操作数从堆栈中,创建与操作在顶部的节点,并挂在它下面的操作数,推新节点到堆栈有关数据:
在推一个压入堆栈 2推入堆栈 在弹出2和A,创建^ -node(A和2所示),将其推堆栈上 在堆栈推2 按A对堆栈 在弹出一个和2相结合,形成了* -node 等 等 Push A onto the stack Push 2 onto the stack Pop 2 and A, create ^-node (with A and 2 below), push it on the stack Push 2 on stack Push A on stack Pop A and 2 and combine to form the *-node etc. etc.
下面是一个 LINQPad 程序,可以与尝试:
Here is a LINQPad program that can be experimented with:
// Add the following two using-directives to LINQPad:
// System.Drawing
// System.Drawing.Imaging
static Bitmap _Dummy = new Bitmap(16, 16, PixelFormat.Format24bppRgb);
static Font _Font = new Font("Arial", 12);
void Main()
{
var elementsAsString = "A 2 ^ 2 A * B * - B 2 ^ + A B - / 2 ^";
var elements = elementsAsString.Split(' ');
var stack = new Stack<Node>();
foreach (var element in elements)
if (IsOperator(element))
{
Node rightOperand = stack.Pop();
Node leftOperand = stack.Pop();
stack.Push(new Node(element, leftOperand, rightOperand));
}
else
stack.Push(new Node(element));
Visualize(stack.Pop());
}
void Visualize(Node node)
{
node.ToBitmap().Dump();
}
class Node
{
public Node(string value)
: this(value, null, null)
{
}
public Node(string value, Node left, Node right)
{
Value = value;
Left = left;
Right = right;
}
public string Value;
public Node Left;
public Node Right;
public Bitmap ToBitmap()
{
Size valueSize;
using (Graphics g = Graphics.FromImage(_Dummy))
{
var tempSize = g.MeasureString(Value, _Font);
valueSize = new Size((int)tempSize.Width + 4, (int)tempSize.Height + 4);
}
Bitmap bitmap;
Color valueColor = Color.LightPink;
if (Left == null && Right == null)
{
bitmap = new Bitmap(valueSize.Width, valueSize.Height);
valueColor = Color.LightGreen;
}
else
{
using (var leftBitmap = Left.ToBitmap())
using (var rightBitmap = Right.ToBitmap())
{
int subNodeHeight = Math.Max(leftBitmap.Height, rightBitmap.Height);
bitmap = new Bitmap(
leftBitmap.Width + rightBitmap.Width + valueSize.Width,
valueSize.Height + 32 + subNodeHeight);
using (var g = Graphics.FromImage(bitmap))
{
int baseY = valueSize.Height + 32;
int leftTop = baseY; // + (subNodeHeight - leftBitmap.Height) / 2;
g.DrawImage(leftBitmap, 0, leftTop);
int rightTop = baseY; // + (subNodeHeight - rightBitmap.Height) / 2;
g.DrawImage(rightBitmap, bitmap.Width - rightBitmap.Width, rightTop);
g.DrawLine(Pens.Black, bitmap.Width / 2 - 4, valueSize.Height, leftBitmap.Width / 2, leftTop);
g.DrawLine(Pens.Black, bitmap.Width / 2 + 4, valueSize.Height, bitmap.Width - rightBitmap.Width / 2, rightTop);
}
}
}
using (var g = Graphics.FromImage(bitmap))
{
float x = (bitmap.Width - valueSize.Width) / 2;
using (var b = new SolidBrush(valueColor))
g.FillRectangle(b, x, 0, valueSize.Width - 1, valueSize.Height - 1);
g.DrawRectangle(Pens.Black, x, 0, valueSize.Width - 1, valueSize.Height - 1);
g.DrawString(Value, _Font, Brushes.Black, x + 1, 2);
}
return bitmap;
}
}
bool IsOperator(string s)
{
switch (s)
{
case "*":
case "/":
case "^":
case "+":
case "-":
return true;
default:
return false;
}
}
输出: