### All Computer Science Resources

## Example Questions

### Example Question #31 : Standard Operations & Algorithms

What is the difference between inorder traversal of a binary search tree and a preorder traversal?

**Possible Answers:**

Preorder searches through the tree from lowest to highest, inorder does not.

In order traversal processes the right subtree, then the root node, then the left subtree, whereas preorder processes the left subtree, then the root node, the the right subtree.

They are similar.

The only difference is that in order processes the root node, whereas preorder does not.

In order traversal processes the left subtree, then the root node, then the right subtree, whereas preorder processes the root node, left subtree, then right subtree.

**Correct answer:**

In order traversal processes the left subtree, then the root node, then the right subtree, whereas preorder processes the root node, left subtree, then right subtree.

In this case, the names help to identify the different types of traversals. In order traversal processes the binary tree "in order", meaning it will go through the left subtree, process the node, then go on to the right. Why is this in order? Because remember, a binary sort tree lists any value less than the node to the left, and any value greater to the right, so in order traversal will actually go from greatest to smallest.

Preorder traversal is the same except that it processes the root node **first,** hence the "pre" order.

### Example Question #2 : Binary

public static int foo(int[] arr, int x) {

int a = 0;

int b = arr.length - 1;

while(b >= a) {

int c = (a + b) / 2;

int v = arr[c];

if(v == x) {

return c;

} else if(v < x) {

a = c + 1;

} else {

b = c - 1;

}

}

return -1;

}

What is the value of *y* in the code below:

int[] vals = {1,3,4,5,6,31,41,51};

int x = 41;

int y = foo(vals,41);

**Possible Answers:**

7

-1

41

5

6

**Correct answer:**

6

The first thing to notice in the code for the method *foo* is that it implements the algorithm for a binary search. The value *c* functions as the midpoint for the algorithm. This requires that the list be in order. The values *a* and *b* are used to control the loop so as to let you search the correct portions of the list. If the value in the middle of the list is not equal to what you are looking for (i.e. *x*), then it looks either "above" or "below" in the list (since it is presumed to be in order). This is what is happening when either *a* or *b *is changed.

So, this searches for the value and returns the *index number* for that value if it is found. Otherwise, it returns -1. Since the value 41 is in the array, it returns 6.

If you do not recognize this algorithm, you are likely to have some problems—and will have to trace the code!!

### Example Question #3 : Binary

Which of the following is true for a binary search algorithm?

**Possible Answers:**

The array must be converted into a binary search tree before searching for the value.

The array being searched must be sorted before searching.

The array can be unsorted but will run slower than if it is sorted.

The array needs extra space in order to accomodate swap operations that are part of the search.

The array must be probed in order from its beginning to its end.

**Correct answer:**

The array being searched must be sorted before searching.

The binary search presumes that the array being searched is already sorted. This is because of how it probes the values in the array. It begins at the center of the array to see if the values match. If the value is smaller than this center element, it then presumes that the value is in the lower half of the array. Otherwise, it presumes that the value must be larger and hence in the upper half of the array. It then probes the center of whichever half it has chosen, continuing to do this until the value has certainly not been found.