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.
<span>Which is not a component of a database that describes how data is stored?</span>
Answer:
#include <iostream>
#include <time.h>
#include <string>
using namespace std;
int main(){
srand(time(NULL));
cout<<"Throw dice"<<endl;
int b =0;
int a=0;
a=rand()%6;
b=rand()%6;
for (int i =0;i<1;i++)
{cout<<"dice one: "<<a<<endl;}
for (int i =0;i<1;i++)
{cout<<"dice two: "<<b<<endl;}
if(a>b)
{cout<<"first dice won"<<endl;}
if(b>a)
{cout<<"second dice won"<<endl;}
else{cout<<"they are same"<<endl;
return main();
}
return 0;
}
Explanation:
/*maybe it help you it is almost done*/
The answer is:
Left
The backspace key only deletes whatever way the point is pointing to. In this case, it is to the left.
Correct me if i'm wrong.
Answer:
C is the correct answer to your question
