我最近一直在研究java中的树。我在sanfoundry.com上发现这个代码对表达式树来说非常棒。它接受一个前缀然后打印出前缀表达式作为中缀和后缀的外观,然后最终输出答案。我的问题是我试图找出如何简化它只是采取后缀并打印出答案。因此,它不是读取前缀并完成所有操作,而是只读取后缀并打印出答案。以下是我找到的代码。这是一个简单的修复,只是让它做后缀吗?或者更难的事情?
private void YourRadGridView_AddingNewDataItem(object sender,
Telerik.Windows.Controls.GridView.GridViewAddingNewEventArgs e)
{
YourObjectModel newItem = new YourObjectModel();
newItem.Date = DateTime.Now;
e.NewObject = newItem;
}
这是主要方法。
class ExpressionTree
{
/** class TreeNode **/
class TreeNode
{
char data;
TreeNode left, right;
/** constructor **/
public TreeNode(char data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
/** class StackNode **/
class StackNode
{
TreeNode treeNode;
StackNode next;
/** constructor **/
public StackNode(TreeNode treeNode)
{
this.treeNode = treeNode;
next = null;
}
}
private static StackNode top;
/** constructor **/
public ExpressionTree()
{
top = null;
}
/** function to clear tree **/
public void clear()
{
top = null;
}
/** function to push a node **/
private void push(TreeNode ptr)
{
if (top == null)
top = new StackNode(ptr);
else
{
StackNode nptr = new StackNode(ptr);
nptr.next = top;
top = nptr;
}
}
/** function to pop a node **/
private TreeNode pop()
{
if (top == null)
throw new RuntimeException("Underflow");
else
{
TreeNode ptr = top.treeNode;
top = top.next;
return ptr;
}
}
/** function to get top node **/
private TreeNode peek()
{
return top.treeNode;
}
/** function to insert character **/
private void insert(char val)
{
try
{
if (isDigit(val))
{
TreeNode nptr = new TreeNode(val);
push(nptr);
}
else if (isOperator(val))
{
TreeNode nptr = new TreeNode(val);
nptr.left = pop();
nptr.right = pop();
push(nptr);
}
}
catch (Exception e)
{
System.out.println("Invalid Expression");
}
}
/** function to check if digit **/
private boolean isDigit(char ch)
{
return ch >= '0' && ch <= '9';
}
/** function to check if operator **/
private boolean isOperator(char ch)
{
return ch == '+' || ch == '-' || ch == '*' || ch == '/';
}
/** function to convert character to digit **/
private int toDigit(char ch)
{
return ch - '0';
}
/** function to build tree from input */
public void buildTree(String eqn)
{
for (int i = eqn.length() - 1; i >= 0; i--)
insert(eqn.charAt(i));
}
/** function to evaluate tree */
public double evaluate()
{
return evaluate(peek());
}
/** function to evaluate tree */
public double evaluate(TreeNode ptr)
{
if (ptr.left == null && ptr.right == null)
return toDigit(ptr.data);
else
{
double result = 0.0;
double left = evaluate(ptr.left);
double right = evaluate(ptr.right);
char operator = ptr.data;
switch (operator)
{
case '+' : result = left + right; break;
case '-' : result = left - right; break;
case '*' : result = left * right; break;
case '/' : result = left / right; break;
default : result = left + right; break;
}
return result;
}
}
/** function to get postfix expression */
public void postfix()
{
postOrder(peek());
}
/** post order traversal */
private void postOrder(TreeNode ptr)
{
if (ptr != null)
{
postOrder(ptr.left);
postOrder(ptr.right);
System.out.print(ptr.data);
}
}
/** function to get infix expression */
public void infix()
{
inOrder(peek());
}
/** in order traversal */
private void inOrder(TreeNode ptr)
{
if (ptr != null)
{
inOrder(ptr.left);
System.out.print(ptr.data);
inOrder(ptr.right);
}
}
/** function to get prefix expression */
public void prefix()
{
preOrder(peek());
}
/** pre order traversal */
private void preOrder(TreeNode ptr)
{
if (ptr != null)
{
System.out.print(ptr.data);
preOrder(ptr.left);
preOrder(ptr.right);
}
}
}
答案 0 :(得分:0)
Postfix是一种线性表示法。你根本不需要一棵树来评估它,只是一堆。你是从错误的地方开始的。