java - 二叉树和 botkey troube - 无法遍历树

Question

我正在做一个项目，它与二叉树和植物 key 有关，现在我需要做的是将机器人 key 加载到程序中，然后工作能够按照我的方式按顺序到达每个节点确定树的类型，当我到达叶节点时，它将显示树的类型。现在我在移动它的部分时遇到了麻烦，我不知道如何让它工作，我尝试了很多不同的事情，包括预遍历，我只是想知道是否有人可以帮我把它从可以这么说。我需要编辑的程序部分的代码如下，“chooseOption”方法是用于遍历树的方法。 提前致谢。

/* * 要更改此模板，请选择“工具”|“模板 * 并在编辑器中打开模板。 */

import java.io.FileInputStream;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.Stack;

/**
*
* @author Alex Drinkwater 10077473
*/
public class BotKey implements IBotKey {

private Node root;
private Node currentNode;
// This stack will come in handy when you implement the "undo" facility
private Stack<Node> choiceStack = new Stack<Node>();

@Override
public boolean isLoaded() {
    return root != null;
}

@Override
public int getNbrChildren() {
    if (currentNode == null) {
        return 0;
    } else {
        return currentNode.getNbrChildren();
    }
}

@Override
public String getLeadStatement(int i) {
    return currentNode.getChildStatement(i);
}

@Override
public void chooseOption(int i) {
    //You need to implement this method

    if (root != null) {
        root.getChildNode(i);
        root.getChildStatement(i);

    }


}

@Override
public void undoChoice() {
    //You need to implement this method
}

@Override
public String getResults() {
    //Dummy implementation. You need to supply a proper one
    if (currentNode.isLeaf() == true) {
        currentNode.getName();
    }
    else{
        String s =currentNode + "";
        if(currentNode!= null){
        }
    }
    return "s";

}


   @Override
   public void loadExample1() {
    Node southernLive = new Node("Southern live Oak: Quercus virginiana");
    Node dwarfLive = new Node("Dwarf live oak: Quercus minima");
    Node willow = new Node("Willow oak");
    Node shingle = new Node("Shingle oak");
    Node blackJack = new Node("Blackjack Oak");
    Node northernRed = new Node("Northern red oak");
    Node white = new Node("White oak Quercus alba");
    Node swampChestnut = new Node("Swamp chestnut oak Quercus prinus");


    Node node7 = new Node();
    node7.addChild(white, "Leaves with 5-9 deep lobes");
    node7.addChild(swampChestnut, "Leaves with 21-27 shallow lobes");

    Node node6 = new Node();
    node6.addChild(blackJack, "Leaves mostly with 3 lobes");
    node6.addChild(northernRed, "Leaves mostly with 7-9 lobes");

    Node node5 = new Node();
    node5.addChild(node7,
            "Lobes or teeth rounded or blunt-pointed, no bristles");
    node5.addChild(node6, "Lobes or teeth bristle-tipped");

    Node node4 = new Node();
    node4.addChild(willow, "Leaf narrow, about 4-6 times as long as broad");
    node4.addChild(shingle, "Leaf broad, about 2-3 times as long as broad");

    Node node3 = new Node();
    node3.addChild(dwarfLive, "Mature plant a small shrub");
    node3.addChild(southernLive, "Mature plant a large tree");

    Node node2 = new Node();
    node2.addChild(node4, "Leaves not evergreen");
    node2.addChild(node3, "Leaves evergreen");

    Node node1 = new Node();
    node1.addChild(node5, "Leaves usually with teeth or lobes");
    node1.addChild(node2, "Leaves usually without teeth or lobes");


    root = node1;
    currentNode = root;
}

@Override
public void loadExample2() {
    Node southernLive = new Node("Southern live Oak: Quercus virginiana");
    Node dwarfLive = new Node("Dwarf live oak: Quercus minima");
    Node willow = new Node("Willow oak");
    Node shingle = new Node("Shingle oak");
    Node blackJack = new Node("Blackjack Oak");
    Node northernRed = new Node("Northern red oak");
    Node white = new Node("White oak Quercus alba");
    Node swampChestnut = new Node("Swamp chestnut oak Quercus prinus");


    Node node7 = new Node();
    node7.addChild(white, "Leaves with 5-9 deep lobes");
    node7.addChild(swampChestnut, "Leaves with 21-27 shallow lobes");

    Node node6 = new Node();
    node6.addChild(blackJack, "Leaves mostly with 3 lobes");
    node6.addChild(northernRed, "Leaves mostly with 7-9 lobes");

    Node node5 = new Node();
    node5.addChild(node7,
            "Lobes or teeth rounded or blunt-pointed, no bristles");
    node5.addChild(node6, "Lobes or teeth bristle-tipped");

    Node node4 = new Node();
    node4.addChild(willow, "Leaf narrow, about 4-6 times as long as broad");
    node4.addChild(shingle, "Leaf broad, about 2-3 times as long as broad");

    Node node3 = new Node();
    node3.addChild(dwarfLive, "Mature plant a small shrub");
    node3.addChild(southernLive, "Mature plant a large tree");

    Node node2 = new Node();
    node2.addChild(node4, "Leaves not evergreen");
    node2.addChild(node3, "Leaves evergreen");

    Node node1 = new Node();
    node1.addChild(node5, "Leaves usually with teeth or lobes");
    node1.addChild(node2, "Leaves usually without teeth or lobes");


    root = node1;
    currentNode = root;

    //You need to implement this method
}

@Override
public void readFromFile(FileInputStream fs) {
    //You need to implement this method

    // You may find that a hashmap like this is useful
    HashMap<Integer, Node> map = new HashMap<Integer, Node>();
}

@Override
public boolean canUndo() {
    //Dummy implementation. You need to provide a proper one
    if (choiceStack != null) {
        return true;
    } else {
        return false;
    }
}
}

Answer 1

您有一个植物“数据库”(树形结构)，描述了每种植物，例如:“柳橡树”？那么您想了解搜索时给出了什么？例如，我们有对某种植物的描述:“叶子有 5-9 个深裂片”，我们想要获取该植物可能是的所有植物？在这种情况下，我们需要找到一个包含文本“Leaves with 5-9 deep lobes”的节点。我们称这个节点为startingNode。

然后我们必须从startingNode移动到它的父节点。我们将startingNode 的父节点称为parentNode。然后我们必须获取parentNode 的父节点等等。重复此操作，直到到达具有植物名称的节点。