将中缀字符串转换为二叉树的代码，我收到：无法编译的源代码 - 错误的符号类型

Question

我正在编写一个代码，通过使用 2 个数组列表和一个链表将中缀语句转换为二叉树。我收到一条错误消息：

线程 "main" java.lang.RuntimeException 中的异常：无法编译的源代码 - 错误的符号类型：prog5.InFixToBinaryTreeConverter.precedence 在 prog5.InFixToBinaryTreeConverter.createBinaryTree(InFixToBinaryTreeConverter.java:56) 在 prog5.InFixToBinaryTreeConverter.run(InFixToBinaryTreeConverter.java:31) 在 prog5.Prog5.main(Prog5.java:13)

我很茫然，我该如何解决这些错误？

我的主要方法：

public class Prog5 {

    public static void main(String[] args) {
        InFixToBinaryTreeConverter fp = new InFixToBinaryTreeConverter();
        fp.run("( ( 6 + 2 ) - 5 ) * 8 / 2");
    }  
}

节点class:

public class Node<String> {
    protected String element;
    protected Node<String> left;
    protected Node<String> right;
    int x; 

    public Node(String e, Node left, Node right){
        element = e; //data = element
        this.left = this.right = null;
    }

    public void displayNode(){
        System.out.print(element);
    }
}

二叉树转换器的中缀class:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import static jdk.nashorn.internal.runtime.JSType.isNumber;


public class InFixToBinaryTreeConverter{

    List<String> stack = new ArrayList<>(); //stack
    List<String> inFix= new LinkedList<>(); //queue
    List<Node> btstack = new ArrayList<>(); //stack
    Node root = null;

    //create a no-arg consutrctor that initializes the inFix, stack , & 
    btstack lists

    String expression;



    public void run(String s){ // run method is driver for program
        this.expression = s;
        createInFix();
        createBinaryTree();
    }

    public void createInFix(){
        String[] temporary = expression.split("\s+"); //temp = a

        for (int i = 0; i < temporary.length; i++ ){
            inFix.add(temporary[i]);
        }
    }

    public Node createBinaryTree(){
        this.stack.add("(");
        this.inFix.add(")");

        while(!this.inFix.isEmpty()){
            String variable = this.inFix.remove(0);
            if(isNumber(variable)){
                Node nodeNew = new Node(variable);
                this.btstack.add(nodeNew);
            }
            if(isLeftParentheses(variable)){
                this.stack.add(variable);
            }
            if(isOperator(variable)){
                while(precedence(this.stack.get((this.stack.size()-1, variable)))){
                    Node right = this.btstack.remove((this.btstack.size()-1));
                    Node left = this.btstack.remove((this.btstack.size()-1));
                    Node nodeNew = new Node(this.stack.remove(this.stack.size()-1), left, right);
                    this.btstack.add(nodeNew);
                }
            }
            this.stack.add(variable);
        }
        if(isRightParentheses(variable)){
            if(this.stack.get(this.stack.size()-1) != null){
                while(!isLeftParentheses(this.stack.get(this.stack.size()-1))){
                    Node right = this.btstack.remove((this.btstack.size()-1));
                    Node left = this.btstack.remove((this.btstack.size()-1));
                    Node nodeNew = new Node(this.stack.remove(this.stack.size()-1), left, right);
                    this.btstack.add(nodeNew);
                }
            }
            root = this.btstack.get((this.btstack.size()-1));
        }
        return root;
    }

       void process(Node node){
           System.out.print(node.element+ " ");
       }
        /* Given a binary tree, print its nodes in inorder*/
        private void printInorder(Node node){
            if (node != null){
                printInorder(node.left); // first recur on left child 
                process(node); // then print the data of node 
                printInorder(node.right); // now recur on right child 
            }
        }

        /* Given a binary tree, print its nodes in preorder*/
        private void printPreorder(Node node){
            if (node != null){
                process(node); // first print data of node 
                printPreorder(node.left); // then recur on left sutree 
                printPreorder(node.right); // now recur on right subtree 
            }
        }
        private void printPostorder(Node node){
            if (node != null){
                printPreorder(node.left); // then recur on left sutree 
                printPreorder(node.right); // now recur on right subtree
                process(node); // first print data of node 

            }
        }

        void printPostorder(){ 
            printPostorder(root);  
        }
        void printInorder(){
            printInorder(root);   
        }
        void printPreorder(){    
            printPreorder(root);  
        }

        private static boolean isOperator(String str){ //check to see if c is an operator
            if("+".equals(str) || "-".equals(str) || "*".equals(str) || "/".equals(str) || "^".equals(str)){
                return true;
            }else {
                return false;
            }
        }

        private static boolean precendence(String operator1, String operator2){
            if("^".equals(operator1)){
                return true;
            } else if ("^".equals(operator2)){
                return false;
            }else if ("*".equals(operator1) || "/".equals(operator1)){
                return true;
            }else if("+".equals(operator1) || "-".equals(operator1)){
                if("*".equals(operator2) || "/".equals(operator2)){
                    return false;
                }else{
                    return true;
                }
            }return false;
        }

    private boolean isLeftParentheses(String x) {

        if("(".equals(x)){
            return true;
        }
        else {
            return false;
        }
    }
    private boolean isRightParentheses(String x) {

        if(")".equals(x)){
            return true;
        }
        else {
            return false;
        }
    }
}

Answer 1

您在 while 循环条件中拼写错误 precedence，您在下面定义的方法名为 precendence。

我不确定你使用的是什么编辑器，但在 Eclipse 中这会显示为错误，当你将鼠标悬停在它上面时，它会说该方法不存在。

而 this.stack.get 方法只接受一个 int 参数，但您传递的是 (this.stack.size()-1, variable)。你的意思可能是这样的：

precedence(this.stack.get(this.stack.size() - 1), variable)

将中缀字符串转换为二叉树的代码，我收到：无法编译的源代码 - 错误的符号类型

Code to convert an infix string to a binary tree and I received: Uncompilable source code - Erroneous sym type

java

binary-tree

compiler-errors

infix-notation

nodes