The length of greet is 6 characters including the space at the end. 

greet.substring(0, (len/2)) is equal to greet.substring(0, 3) 

The substring of greet from the zeroth position to second position, not to the third position.

The main thing to look at for this question is the main loop for the code.  This loop first goes through each of the characters in the String s:

int i = 0; i < s.length(); i++

Next, notice the condition on the inner loop:

vals[i % vals.length]

The modulus on i will yield values that are between 0 and 2 (given the length of vals).  This means that you will loop in the inner loop the following sequence of times:

5,4,2,5,4,2,5,4

This will replicate the given letter at the index (i) in the initial string, using 5,4,2,5, etc as the replication count.  Thus, you will replicate "H" 5 times, "e" 4, etc.  This gives you an output:

HHHHHeeeerrvvvvvaaaaeeuuuuussss

It is easiest to think of the values array as a String: "I love you!!".

Now, the loop is going to run for   or  times.  Notice what it does on each iteration.  It swaps the values at i and values.length - i -1.  Thus, it will do the following swaps:

0 and 11

1 and 10

etc...

This sequence of swaps will eventually reverse the array.  Thus, your output is:

!!uoy evol I

Given that strings cannot be internally modified, you will have to store your result in a new string, namely s2.  Now, for each character in s, you will have to make that charater lower case to begin with:

char c = Character.toLowerCase(s.charAt(i));

Next, for the odd values of i, you will need to make your value to be upper case.  The modulus operator is great for this!  You can use % 2 to find the odd values.  When the remainder of a division by 2 is equal to 1, you know you have an odd value.  Hence, you have the condition:

 if(i % 2 == 1) {...

Then, once you appropriately capitalize, you can place your character on s2.

The easiest way to consider this is by commenting on the correct answer.  You must begin by defining the character array:

char[] vals = {'A','t', ' ', '6',' ','a','m','!'};

Next, you must initialize the string value s to be an empty string.  This is critical.  Otherwise, you can't build your string!

String s = "";

Next, you have the loop.  This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.

The only major issue with this code is the use of the == operator to compare the two strings.  Since the user has input this value, will not have an equality on this test.  You must use the method .equals in order to check whether two strings are equal.  The code should be:

if(books[i].equals(userInput)){

...

(There are some cases in which == will work for string comparison, namely when literals are involved.  However, you should avoid relying on this and always use .equals().)

The compareTo method for strings compares to string objects and returns:

• 0 when they are equal
• Something positive when the given string is alphabetically (really, lexicographically) later than the argument to the method.
• Something negative when the given string is alphabetically (really, lexicographically) before the argument to the method.

Our strings could be put in this order:

"Cursus Philosophicus Thomisticus"

"Happy words for happy people"

"Insane words for an insane world"

"Kritik der reinen Vernunft"

"Logica sive Ars Rationalis"

For each of these, we are asking, "Is it later in order than 'Kittens in a cart'?"  This is true for "Kritik der reinen Vernunft" and "Logica sive Ars Rationalis".  Thus, our output is:

[Logica sive Ars Rationalis, Kritik der reinen Vernunft]

Notice the order—this is due to the order in the original array.

The length of greet is 6 characters including the space at the end. 

greet.substring(0, (len/2)) is equal to greet.substring(0, 3) 

The substring of greet from the zeroth position to second position, not to the third position.

A standard Binary Tree inserts into the root first.  It then tries to insert to the "left" for values that are smaller and to the "right" for values that are larger.  Therefore, for the data given, we have the first step:

Next, you will insert "Henrietta" to the left, for that is less than "Isaac":

Next, "Nigel" is greater than "Isaac":

Then, "Buford" is less than "Isaac" and then less than "Henrietta":

This continues through the following stages:

Thus, the last image is your final tree!

When doing a post-order traversal we go in the following order:

left, right, root.

This means that we are doing any left nodes first, then the right nodes, and LASTLY, the root nodes. If a node is a parent, then we must go throught the left and right children first. Since we're doing post-order traversal, the main root is going to be LAST.

In our example, 1 is a parent so we go to it's left child who is also a parent to node 4. This means that 4 is our first number in the traversal. 

POST ORDER TRAVERSAL: 4

Since node 2 doesn't have a right child, we then make that node our next number in the traversal since node 2 it's the left child of node 1.

POST ORDER TRAVERSAL: 4, 2

By now, we have traversed the left nodes of the root. Now we move on to the right subtrees of node 1. Since node 3 is a parent node, we go to it's left child first (node 6). Since node 6 is also a parent, we move on to its children. Node 6 doesn't have a left child. Therefore our next number in the traversal is its right child (node 8) and then the subtree's root (node 6).

POST ORDER TRAVERSAL: 4, 2, 8, 6

Now, we've traversed the left children of node 3 we need to traverse the right child (node 7) who doesn't have any children of its own. 

POST ORDER TRAVERSAL: 4, 2, 8, 6, 7

Lastly since we traversed it's right child, we move to the parent and traverse node 3 and our main root (node 1).

POST ORDER TRAVERSAL: 4, 2, 8, 6, 7, 3, 1