这是我的代码。但是,每当我输入诸如“2 * 3 + 6/3”之类的表达式时,它会将输出打印为“36/3 + 2 *”,这不是正确的后缀格式。有人在我的代码中看到任何错误吗?非常感谢!
package expressiontree;
import java.util.LinkedList;
import java.util.List;
class Stack
{
private List stack = new LinkedList();
public void push(TreeNode node)
{
stack.add(node);
}
public TreeNode pop()
{
int top = stack.size()-1;
if (top < 0) return null;
TreeNode node = (TreeNode)stack.get(top);
stack.remove(top);
return node;
}
}
class TreeNode
{
TreeNode left;
TreeNode right;
char data;
TreeNode(){}
TreeNode(char c)
{
this.data = c;
}
}
public class ExpressionTree {
String expression;
TreeNode tree;
Stack operandstack = new Stack();
Stack operatorstack = new Stack();
private boolean isOperator(char c)
{
String operator = "+-*/";
if(operator.indexOf(c)==-1)
return false;
return true;
}
// parses an expression and creates two stacks
private void parseExpression(String exp)
{
for(int i=0; i < exp.length(); i++) {
char c = exp.charAt(i);
if(isOperator(c))
operatorstack.push(new TreeNode(c));
else{
operandstack.push(new TreeNode(c));
}
}
}
private TreeNode create_tree()
{
TreeNode node = new TreeNode();
while((node = operatorstack.pop())!=null)
{
TreeNode oper1 = operandstack.pop();
TreeNode oper2 = operandstack.pop();
TreeNode exp = create_expression(oper1, node, oper2);
operandstack.push(exp);
}
return operandstack.pop();
}
private TreeNode create_expression(TreeNode oper1, TreeNode operator, TreeNode oper2)
{
TreeNode expression = new TreeNode(operator.data);
expression.left = oper1;
expression.right = oper2;
return expression;
}
public static void postfix(TreeNode tn){
if (tn == null){
return;
}
else{
postfix(tn.left);
postfix(tn.right);
System.out.print(tn.data);
}
}
public static void main(String[] args)
{
ExpressionTree et = new ExpressionTree();
String expression = "2*3+6/3";
et.parseExpression(expression);
TreeNode tn = et.create_tree();
postfix(tn);
System.out.println();
}
}
答案 0 :(得分:1)
我可以看到代码的两个基本问题。
1)在parseExpression
中,操作数和运算符被盲目地推入各自的堆栈中。您需要注意运算符优先级(即*和/的优先级高于+ - )。为了处理运算符优先级,基本上当您看到优先级低于或等于当前运算符堆栈的运算符时,必须先评估堆栈中当前的内容,然后再将操作符推入堆栈。
2)在create_tree
中,从堆栈中分配第一个和第二个操作数的代码使用第一个pop作为第一个操作数,第二个pop作为第二个操作数。它应该是相反的方式(即第一个pop是第二个操作数,第二个pop是第一个操作数)
尝试这些更改,看看它是否能解决您的问题。