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
- LinkedBinaryTreeTest
- BinaryTree
- DecisionTree
- Expression Tree
- 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% |
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