20162314 Experiment 2 - Tree

Experiment report of Besti

course：《Program Design & Data Structures》

Class： 1623

Student Name： Wang, Yixiao

Student Number：20162314

Tutor：Mr.Lou、Mr.Wang

Experiment date：2017.10.27

Secret level： Unsecretive

Experiment time：60 minutes

Major/Elective：Major

Experiment order：6

Experiment content

1. LinkedBinaryTreeTest
   
   

1. BinaryTree
   
   

1. DecisionTree
   
   

1. Expression Tree
   
   

1. BinarySearchTree
   
   

6.Analyze Source Code (Red-Black Tree)

Experiment situation

Exp1 LinkedBinaryTreeTest

• It's easy to finish this experiment.
• To start with , import to form a new LinkedBinaryTree to start the experiment.
• Next , use the method GetRight GetLeft to ealuation.
• Then , new element "ABC".
• Last , Assert.assertequals();

Exp2 BinaryTree

• To start with , use the number to replace "A B C D ....."
• Next , new newNode
• Then, creat BinTree to get left child and right child
• Use three methods preOrderTraverse,inOrderTraverse,postOrderTraverse.

Exp3 Decision Tree

• To start with , creat a class TwentyQuestionsPlayer
• String the Questions and the answers as the order
• New them
• Write a method play(),Yes =Y , No=N .
• Next , write a function to finish this experiment.

Exp4 ExpressionTree

Exp5 BinarySearchTreeTest

• To start with , creat a class BinarySearchTree
• public method Node find, insert,preorder,inorder,postorder,getMinNode get MaxNode，Delete.
• creat three new Nodes(id and name);
• Node n1 n2 n3 n4 ...n7 n8
• Use method insert to n4 n5 n6 n7 n8
• Bst.inorder and Systemin.

Exp6 Red-Black Tree (Analyze Source Code)

• TreeMap底层通过红黑树（Red-Black tree）实现，也就意味着containsKey(), get(), put(), remove()都有着log(n)的时间复杂度。
  
  

• SortedMap m = Collections.synchronizedSortedMap(new TreeMap(...));

• 结构调整过程包含两个基本操作：左旋（Rotate Left），右旋（RotateRight）。
  
  

• Rotate Left

  //Rotate Left
private void rotateLeft(Entry<K,V> p) {
    if (p != null) {
        Entry<K,V> r = p.right;
        p.right = r.left;
        if (r.left != null)
            r.left.parent = p;
        r.parent = p.parent;
        if (p.parent == null)
            root = r;
        else if (p.parent.left == p)
            p.parent.left = r;
        else
            p.parent.right = r;
        r.left = p;
        p.parent = r;
    }
}

• Rotate Right

  //Rotate Right
private void rotateRight(Entry<K,V> p) {
    if (p != null) {
        Entry<K,V> l = p.left;
        p.left = l.right;
        if (l.right != null) l.right.parent = p;
        l.parent = p.parent;
        if (p.parent == null)
            root = l;
        else if (p.parent.right == p)
            p.parent.right = l;
        else p.parent.left = l;
        l.right = p;
        p.parent = l;
    }
}

• .get()
  
  

  //getEntry()方法
final Entry<K,V> getEntry(Object key) {
    ......
    if (key == null)//不允许key值为null
        throw new NullPointerException();
    Comparable<? super K> k = (Comparable<? super K>) key;//使用元素的自然顺序
    Entry<K,V> p = root;
    while (p != null) {
        int cmp = k.compareTo(p.key);
        if (cmp < 0)//向左找
            p = p.left;
        else if (cmp > 0)//向右找
            p = p.right;
        else
            return p;
    }
    return null;
}

• .put()

  public V put(K key, V value) {
    ......
    int cmp;
    Entry<K,V> parent;
    if (key == null)
        throw new NullPointerException();
    Comparable<? super K> k = (Comparable<? super K>) key;//使用元素的自然顺序
    do {
        parent = t;
        cmp = k.compareTo(t.key);
        if (cmp < 0) t = t.left;//向左找
        else if (cmp > 0) t = t.right;//向右找
        else return t.setValue(value);
    } while (t != null);
    Entry<K,V> e = new Entry<>(key, value, parent);//创建并插入新的entry
    if (cmp < 0) parent.left = e;
    else parent.right = e;
    fixAfterInsertion(e);//调整
    size++;
    return null;
}

  //红黑树调整函数fixAfterInsertion()
private void fixAfterInsertion(Entry<K,V> x) {
    x.color = RED;
    while (x != null && x != root && x.parent.color == RED) {
        if (parentOf(x) == leftOf(parentOf(parentOf(x)))) {
            Entry<K,V> y = rightOf(parentOf(parentOf(x)));
            if (colorOf(y) == RED) {//如果y为null，则视为BLACK
                setColor(parentOf(x), BLACK);              // 情况1
                setColor(y, BLACK);                        // 情况1
                setColor(parentOf(parentOf(x)), RED);      // 情况1
                x = parentOf(parentOf(x));                 // 情况1
            } else {
                if (x == rightOf(parentOf(x))) {
                    x = parentOf(x);                       // 情况2
                    rotateLeft(x);                         // 情况2
                }
                setColor(parentOf(x), BLACK);              // 情况3
                setColor(parentOf(parentOf(x)), RED);      // 情况3
                rotateRight(parentOf(parentOf(x)));        // 情况3
            }
        } else {
            Entry<K,V> y = leftOf(parentOf(parentOf(x)));
            if (colorOf(y) == RED) {
                setColor(parentOf(x), BLACK);              // 情况4
                setColor(y, BLACK);                        // 情况4
                setColor(parentOf(parentOf(x)), RED);      // 情况4
                x = parentOf(parentOf(x));                 // 情况4
            } else {
                if (x == leftOf(parentOf(x))) {
                    x = parentOf(x);                       // 情况5
                    rotateRight(x);                        // 情况5
                }
                setColor(parentOf(x), BLACK);              // 情况6
                setColor(parentOf(parentOf(x)), RED);      // 情况6
                rotateLeft(parentOf(parentOf(x)));         // 情况6
            }
        }
    }
    root.color = BLACK;
}

Code hosting

PSP5.1(Personal Software Process)

Steps	Time	percent
requirement	45minutes	16.7%
design	50minutes	18.5%
coding	1.5hours	32.2%
test	30minutes	11.1%
summary	55minutes	19.2%