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nata0808 [166]
3 years ago
7

Wired network are the most reliable and provide the highest speed?

Computers and Technology
1 answer:
lbvjy [14]3 years ago
8 0

A wired connection is almost always reliable and fastest, so that one gets a True.

When it comes to wireless networks, it really just depends on how you're setting them up.  A normal wireless set up is generally effortless to set up because a lot of router manufacturers include wizards in the router's firmware to help you get started, so that gets a False.

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Answer:

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Explanation:

// Class for BinarySearchTreeNode

class TreeNode

{

// To store key value

public int key;

// To point to left child

public TreeNode left;

// To point to right child

public TreeNode right;

// Default constructor to initialize instance variables

public TreeNode(int key)

{

this.key = key;

left = null;

right = null;

key = 0;

}// End of default constructor

}// End of class

// Defines a class to crate binary tree

class BinaryTree

{

// Creates root node

TreeNode root;

int numberElement;

// Default constructor to initialize root

public BinaryTree()

{

this.root = null;

numberElement = 0;

}// End of default constructor

// Method to insert key

public void insert(int key)

{

// Creates a node using parameterized constructor

TreeNode newNode = new TreeNode(key);

numberElement++;

// Checks if root is null then this node is the first node

if(root == null)

{

// root is pointing to new node

root = newNode;

return;

}// End of if condition

// Otherwise at least one node available

// Declares current node points to root

TreeNode currentNode = root;

// Declares parent node assigns null

TreeNode parentNode = null;

// Loops till node inserted

while(true)

{

// Parent node points to current node

parentNode = currentNode;

// Checks if parameter key is less than the current node key

if(key < currentNode.key)

{

// Current node points to current node left

currentNode = currentNode.left;

// Checks if current node is null

if(currentNode == null)

{

// Parent node left points to new node

parentNode.left = newNode;

return;

}// End of inner if condition

}// End of outer if condition

// Otherwise parameter key is greater than the current node key

else

{

// Current node points to current node right

currentNode = currentNode.right;

// Checks if current node is null

if(currentNode == null)

{

// Parent node right points to new node

parentNode.right = newNode;

return;

}// End of inner if condition

}// End of outer if condition

}// End of while

}// End of method

// Method to check tree is balanced or not

private int checkBalance(TreeNode currentNode)

{

// Checks if current node is null then return 0 for balanced

if (currentNode == null)

return 0;

// Recursively calls the method with left child and

// stores the return value as height of left sub tree

int leftSubtreeHeight = checkBalance(currentNode.left);

// Checks if left sub tree height is -1 then return -1

// for not balanced

if (leftSubtreeHeight == -1)

return -1;

// Recursively calls the method with right child and

// stores the return value as height of right sub tree

int rightSubtreeHeight = checkBalance(currentNode.right);

// Checks if right sub tree height is -1 then return -1

// for not balanced

if (rightSubtreeHeight == -1) return -1;

// Checks if left and right sub tree difference is greater than 1

// then return -1 for not balanced

if (Math.abs(leftSubtreeHeight - rightSubtreeHeight) > 1)

return -1;

// Returns the maximum value of left and right subtree plus one

return (Math.max(leftSubtreeHeight, rightSubtreeHeight) + 1);

}// End of method

// Method to calls the check balance method

// returns false for not balanced if check balance method returns -1

// otherwise return true for balanced

public boolean balanceCheck()

{

// Calls the method to check balance

// Returns false for not balanced if method returns -1

if (checkBalance(root) == -1)

return false;

// Otherwise returns true

return true;

}//End of method

// Method for In Order traversal

public void inorder()

{

inorder(root);

}//End of method

// Method for In Order traversal recursively

private void inorder(TreeNode root)

{

// Checks if root is not null

if (root != null)

{

// Recursively calls the method with left child

inorder(root.left);

// Displays current node value

System.out.print(root.key + " ");

// Recursively calls the method with right child

inorder(root.right);

}// End of if condition

}// End of method

}// End of class BinaryTree

// Driver class definition

class BalancedBinaryTreeCheck

{

// main method definition

public static void main(String args[])

{

// Creates an object of class BinaryTree

BinaryTree treeOne = new BinaryTree();

// Calls the method to insert node

treeOne.insert(1);

treeOne.insert(2);

treeOne.insert(3);

treeOne.insert(4);

treeOne.insert(5);

treeOne.insert(8);

// Calls the method to display in order traversal

System.out.print("\n In order traversal of Tree One: ");

treeOne.inorder();

if (treeOne.balanceCheck())

System.out.println("\n Tree One is balanced");

else

System.out.println("\n Tree One is not balanced");

BinaryTree

BinaryTree treeTwo = new BinaryTree();

treeTwo.insert(10);

treeTwo.insert(18);

treeTwo.insert(8);

treeTwo.insert(14);

treeTwo.insert(25);

treeTwo.insert(9);

treeTwo.insert(5);

System.out.print("\n\n In order traversal of Tree Two: ");

treeTwo.inorder();

if (treeTwo.balanceCheck())

System.out.println("\n Tree Two is balanced");

else

System.out.println("\n Tree Two is not balanced");

}// End of main method

}// End of driver class

5 0
3 years ago
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