Design 3 classes: Computer - Superclass
1 answer:
Answer:
Explanation:
The following code is written in Java. It creates the three classes mentioned and a Tester class that contains the main method. The Computer class contains the memory variable since both laptops and Desktops need memory. The screen size variable is placed separately in the Laptop and Desktop class since desktops may or may not have a monitor so this variable cannot be placed in the Computer class. The batteryLife variable is only in the Laptop class because Desktops do not have batteries. Finally, the monitor variable is only placed in the Desktop class since Laptop's come with built-in monitors. The tester class seen in the picture below tests the creation of both of these objects and it executes without any error.
class Computer {
int memory;
public int getMemory() {
return memory;
}
public void setMemory(int memory) {
this.memory = memory;
}
}
class Laptop extends Computer {
int screenSize;
double batteryLife;
public int getScreenSize() {
return screenSize;
}
public void setScreenSize(int screenSize) {
this.screenSize = screenSize;
}
public double getBatteryLife() {
return batteryLife;
}
public void setBatteryLife(double batteryLife) {
this.batteryLife = batteryLife;
}
}
class Desktop extends Computer {
boolean monitor;
int screenSize;
public boolean isMonitor() {
return monitor;
}
public void setMonitor(boolean monitor) {
this.monitor = monitor;
}
public int getScreenSize() {
return screenSize;
}
public void setScreenSize(int screenSize) {
this.screenSize = screenSize;
}
}
class Tester {
public static void main(String[] args) {
Laptop computer1 = new Laptop();
Desktop computer2 = new Desktop();
}
}
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Answer:
30,33,37,18,23,34,15,38,40,17
To construct the AVL tree, follow these steps and diagrams are shown in the image:
• Add 30 to the tree as the root node. Then add 33 as the right child because of 33 is greater than 30 and AVL tree is a binary search tree.
• Then add 37 as the right child of 33. Here the balance factor of node 30 becomes 0-2 = -2, unbalanced.
Use RR rotation, make node 33 the root node, 30 as the left child of 33 and 37 as the right child of 33.
• Now add 18 as the left child of 30. And 23 as the right child of 18. Here the balance factor of 30 becomes 2-0 = 2. It’s unbalanced.
Use LR rotation, make 23 the parent of 18 and 30.
• Now add 34 as the left child of 37 and 15 as the left child of 18. Add 38 as the right child of 37. And then add 40 as the right child of 38.
• Now adding 17 as the right child of 15 makes the tree unbalanced at 18.
Use LR rotation, make 17 as the parent of 15 and 18.
Explanation:
The balance factor of a node can be either 0,1 or -1. Else the tree is called unbalanced at the node.
If the inserted node is in the left subtree of the left subtree of the unbalance node, then perform LL rotation.
If the inserted node is in the right subtree of the right subtree of the unbalance node, then perform RR rotation.
If the inserted node is in the left subtree of the right subtree of the unbalance node, then perform RL rotation.
If the inserted node is in the right subtree of the left subtree of the unbalance node, then perform LR rotation.
