20192306 2020-2021-1 《数据结构与面向对象程序设计》实验八报告
20192306 2020-2021-1 《数据结构与面向对象程序设计》实验八报告
课程：《程序设计与数据结构》
班级： 1923
姓名： 孙洪丽
学号： 20192306
实验教师：王志强
实验日期：2020年12月3日
必修/选修： 必修
1.实验内容
- 参考教材PP16.1,完成链树LinkedBinaryTree的实现（getRight,contains,toString,preorder,postorder）
- 用JUnit或自己编写驱动类对自己实现的LinkedBinaryTree进行测试，提交测试代码运行截图，要全屏，包含自己的学号信息
- 课下把代码推送到代码托管平台
- 基于LinkedBinaryTree，实现基于（中序，先序）序列构造唯一一棵二㕚树的功能，比如给出中序HDIBEMJNAFCKGL和后序ABDHIEJMNCFGKL,构造出附图中的树
- 用JUnit或自己编写驱动类对自己实现的功能进行测试，提交测试代码运行截图，要全屏，包含自己的学号信息
- 课下把代码推送到代码托管平台
- 自己设计并实现一颗决策树
- 提交测试代码运行截图，要全屏，包含自己的学号信息
- 课下把代码推送到代码托管平台
- 输入中缀表达式，使用树将中缀表达式转换为后缀表达式，并输出后缀表达式和计算结果（如果没有用树，正常评分。如果用到了树，即使有小的问题，也酌情给满分）
- 提交测试代码运行截图，要全屏，包含自己的学号信息
2.实验过程及结果
完成链树LinkedBinaryTree的实现
- BTNode
import java.util.ArrayList;
public class BTNode<T>
{
protected T element;
protected BTNode<T> left, right;
public BTNode (T element)
{
this.element = element;
left = right = null;
}
public T getElement()
{
return element;
}
public void setElement (T element)
{
this.element = element;
}
public BTNode<T> getLeft()
{
return left;
}
public void setLeft (BTNode<T> left)
{
this.left = left;
}
public BTNode<T> getRight()
{
return right;
}
public void setRight (BTNode<T> right)
{
this.right = right;
}
public BTNode<T> find (T target) {
BTNode<T> result = null;
if (element.equals(target))
result = this;
else
{
if (left != null)
result = left.find(target);
if (result == null && right != null)
result = right.find(target);
}
return result;
}
public int count() {
int result = 1;
if (left != null)
result += left.count();
if (right != null)
result += right.count);
return result;
}
public void inorder (ArrayList<T> iter) {
if (left != null)
left.inorder (iter);
iter.add (element);
if (right != null)
right.inorder (iter);
}
public void preorder (ArrayList<T> iter) {
iter.add (element);
if (left != null)
left.inorder (iter);
if (right != null)
right.inorder (iter);
}
public void postorder (ArrayList<T> iter) {
if (left != null)
left.inorder (iter);
if (right != null)
right.inorder (iter);
iter.add (element);
}
public char print() {
return (char) element;
}
}
- LinkedBinaryTree
import java.util.Iterator;
public class LinkedBinaryTree<T> implements BinaryTree<T> {
public BTNode<T> root;//定义根结点
public BTNode left;
public BTNode right;
public LinkedBinaryTree()
{
root = null;
}//未赋值时，根结点为空
public LinkedBinaryTree (T element)
{
root = new BTNode<T>(element);
}//进行根结点赋值
public LinkedBinaryTree (T element, LinkedBinaryTree<T> left,
LinkedBinaryTree<T> right)//根、左、右、再思考
{
root = new BTNode<T>(element);
root.setLeft(left.root);
root.setRight(right.root);
}
public T getRootElement() throws Exception {
if (root == null)
throw new Exception ("Get root operation "
+ "failed. The tree is empty.");
return root.getElement();
}
//先左子树
public LinkedBinaryTree<T> getLeft() throws Exception {
if (root == null)
throw new Exception ("Get left operation "
+ "failed. The tree is empty.");
LinkedBinaryTree<T> result = new LinkedBinaryTree<T>();
result.root = root.getLeft();
return result;
}
public T find (T target) throws Exception {
BTNode<T> node = null;
if (root != null)
node = root.find(target);
if (node == null)
throw new Exception("Find operation failed. "
+ "No such element in tree.");
return node.getElement();
}
public int size()
{
int result = 0;
if (root != null)
result = root.count();
return result;
}
public Iterator<T> inorder()
{
ArrayIterator<T> iter = new ArrayIterator<T>();
if (root != null)
root.inorder (iter);
return iter;
}
public Iterator<T> levelorder() throws EmptyCollectionException {
LinkedQueue<BTNode<T>> queue = new LinkedQueue<BTNode<T>>();
ArrayIterator<T> iter = new ArrayIterator<T>();
if (root != null)
{
queue.enqueue(root);
while (!queue.isEmpty())
{
BTNode<T> current = queue.dequeue();
iter.add (current.getElement());
if (current.getLeft() != null)
queue.enqueue(current.getLeft());
if (current.getRight() != null)
queue.enqueue(current.getRight());
}
}
return iter;
}
public Iterator<T> ArrayIterator()
{
return inorder();
}
//再右子树
public LinkedBinaryTree<T> getRight() throws Exception {
if (root == null)
throw new Exception ("Get Right operation "
+ "failed. The tree is empty.");
LinkedBinaryTree<T> result = new LinkedBinaryTree<T>();
result.root = root.getRight();
return result;
}
public boolean contains (T target) throws Exception {
BTNode<T> node = null;
boolean result = true;
if (root != null)
node = root.find(target);
if(node == null)
result = false;
return result;
}
public boolean isEmpty() {
if(root!=null)
return false;
else
return true;
}
public String toString() {
ArrayIterator<T> list = (ArrayIterator<T>) preorder();
String result = "<top of Tree>\n";
for(T i : list){
result += i + "\t";
}
return result + "<bottom of Tree>";
}
public Iterator<T> preorder() {
ArrayIterator<T> list = new ArrayIterator<>();
if(root!=null)
root.preorder(list);
return list;
}
public Iterator<T> postorder() {
ArrayIterator<T> list = new ArrayIterator<>();
if(root!=null)
root.postorder(list);
return list;
}
@Override
public Iterator<T> iterator() {
return null;
}
}
- 结果截图
码云
基于（中序，先序）序列构造唯一一棵二㕚树的功能
- 扩展LinkedBinaryTree
public BTNode construct(char[] pre, char[] in) {
if (pre.length == 0 || in.length == 0) {
return null;
}
BTNode<Character> tree = new BTNode<Character>(pre[0]);
int index = search(0, in.length, in, tree.getElement());
tree.setLeft(construct(Arrays.copyOfRange(pre, 1, index + 1), Arrays.copyOfRange(in, 0, index)));
tree.setRight(construct(Arrays.copyOfRange(pre, index + 1, pre.length),
Arrays.copyOfRange(in, index + 1, in.length)));
return tree;
}
public int search(int start, int end, char[] inOrders, char data) {
for (int i = start; i < end; i++) {
if (data == inOrders[i]) {
return i;
}
}
return -1;
}
public void preOrder(BTNode<T> Tree) {
System.out.print(Tree.getElement() + " ");
BTNode<T> leftTree = Tree.left;
if (leftTree != null) {
preOrder(leftTree);
}
BTNode<T> rightTree = Tree.right;
if (rightTree != null) {
preOrder(rightTree);
}
}
public static void inOrder(BTNode tree) {
if (tree == null)
return;
else {
inOrder(tree.left);
System.out.print(tree.print() + " ");
inOrder(tree.right);
}
}
- 结果截图
码云
自己设计并实现一颗决策树
- Decision
import java.util.Scanner;
public class Decision {
private LinkedBinaryTree<String> tree;
public Decision(){
String e1 = "您知道王俊凯吗？";
String e2 = "您知道我国的火星探测任务吗？";
String e3 = "希望您多了解他，他很棒。";
String e4 = "您知道它的主题曲《摘星少年》是由谁演唱的吗？";
String e5 = "那没事了，希望您多多关注国家大事纪。";
String e6 = "太好了，很高兴和您分享他。";
String e7 = "是由王俊凯演唱的，希望您多多关注。";
LinkedBinaryTree<String> n2,n3,n4,n5,n6,n7;
n3 = new LinkedBinaryTree<String>(e3);
n5 = new LinkedBinaryTree<String>(e5);
n6 = new LinkedBinaryTree<String>(e6);
n7 = new LinkedBinaryTree<String>(e7);
n4 = new LinkedBinaryTree<String>(e4,n6,n7);
n2 = new LinkedBinaryTree<String>(e2,n4,n5);
tree = new LinkedBinaryTree<String>(e1,n2,n3);
}
public void diagose() throws Exception {
Scanner scan = new Scanner(System.in);
LinkedBinaryTree<String> current = tree;
while(current.size()>1)
{
System.out.println(current.getRootElement());
if(scan.nextLine().equalsIgnoreCase("Y"))
current = current.getLeft();
else
current = current.getRight();
}
System.out.println(current.getRootElement());
}
}
- DecisionTest
public class DecisionTest {
public static void main(String[]args) throws Exception {
Decision a = new Decision();
a.diagose();
}
}
- 结果截图（各种逻辑合理，符合预期，决策树设计成功）
码云
中缀表达式转换
- 普通使用栈设计转换代码
import java.util.Stack;
public class Fix {
static Stack<Character> op = new Stack<>();
public static Float getv(char op, Float f1, Float f2) {
if (op == '+') return f2 + f1;
else if (op == '-') return f2 - f1;
else if (op == '*') return f2 * f1;
else if (op == '/') return f2 / f1;
else return Float.valueOf(-0);
}
public static float calrp(String rp) {
Stack<Float> v = new Stack<>();
char[] arr = rp.toCharArray();
int len = arr.length;
for (int i = 0; i < len; i++) {
Character ch = arr[i];
if (ch >= '0' && ch <= '9') v.push(Float.valueOf(ch - '0'));
else v.push(getv(ch, v.pop(), v.pop()));
}
return v.pop();
}
public static String getrp(String s) {
char[] arr = s.toCharArray();
int len = arr.length;
String out = "";
for (int i = 0; i < len; i++) {
char ch = arr[i];
if (ch == ' ') continue;
if (ch >= '0' && ch <= '9') {
out += ch;
continue;
}
if (ch == '(')
op.push(ch);
if (ch == '+' || ch == '-') {
while (!op.empty() && (op.peek() != '('))
out += op.pop();
op.push(ch);
continue;
}
if (ch == '*' || ch == '/') {
while (!op.empty() && (op.peek() == '*' || op.peek() == '/'))
out += op.pop();
op.push(ch);
continue;
}
if (ch == ')') {
while (!op.empty() && op.peek() != '(')
out += op.pop();
op.pop();
continue;
}
}
while (!op.empty()) out += op.pop();
return out;
}
}
- 测试
import java.util.Scanner;
public class Test4 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("请输入表达式:");
String s = scan.nextLine();
Fix fix = new Fix();
System.out.println("后缀表达式为:\n"+Fix.getrp(s));
System.out.println("计算结果为:\n"+fix.calrp(Fix.getrp(s)));
}
}
- 结果截图
码云
其他（感悟、思考等）
- 树是一个比较难掌握的数据结构，虽然上学期在离散课上接触过了，但知识掌握程度还停留在理论方面，真正代码实现的时候——特别是许多方法涉及递归算法的时候——我觉得还是很困难的，需要更多的实践去练习。
