Answer:
B.O(n).
Explanation:
Since the time complexity of visiting a node is O(1) in iterative implementation.So the time complexity of visiting every single node in binary tree is O(n).We can use level order traversal of a binary tree using a queue.Which can visit every node in O(n) time.Level order traversal do it in a single loop without doing any extra traversal.
Answer:
It is the ALU or the Arithmetic Logic Unit.
Explanation:
It is the ALU. However, keep in mind that registers and buses do a very important task. The number of registers we have, faster is the processing, and the opposite is true as well. And there is a reason behind this if we have different channels for sending and receiving the data from the memory, and several registers for storing the data, and we can formulate the requirement seeing the requirements for full adder and half adders. Remember we need to store several variables in case of the full adder, and which is the carry, and if we have separate registers for each of them, our task becomes easier. Remember its the CU that tells the ALU what operation is required to be performed. Also remember we have the same channel for input and output in the case of Van Neumann architecture, as we have a single bus. and we also have a single shared memory. And Harvard architecture is an advanced version of it.
The steps I would take are 1. Making a strong password with symbols and numbers. 2. I will set up a 2 way authentication ther for if someone would be trying to hack me I would get an alert in order to stop the attempt. 3 I will not use the same password each time.
ARPANET would not carry it
Answer:
//Here is the for loop in C.
for(n=10;n>0;n--)
{
printf("count =%d \n",n);
}
Explanation:
Since C is a procedural programming language.Here if a loop that starts with n=10; It will run till n becomes 0. When n reaches to 0 then loop terminates otherwise it print the count of n.
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{ // variables
int n;
// for loop that runs 10 times
// when n==0 then loop terminates
for(n=10;n>0;n--)
{
cout<<"count ="<<n<<endl;
}
return 0;
}
Output:
count =10
count =9
count =8
count =7
count =6
count =5
count =4
count =3
count =2
count =1
