Please Provide The Code In Java And Please Perform A Trace with num = 39312 preferably a picture!!

R02: Sum Digits Compute the sum of the digits of a number. Note that % 10 will give the rightmost digit and / 10 will remove the rightmost digit. Assume the number will be >= 0. Example Input Expected Output 123 6 3919 22 17 8 0 0 First, let us consider the base case. This might not seem obvious at first, but the description gives a hint as to what we need to do. First, we need to reduce the problem down somehow. Removing a digit sounds like a reduction of some kind so lets see what happens if we remove the rightmost digit. (Note, this is integer division. Decimals have no place here). 123 / 10 = 12 12 / 10 = 1 1 / 10 = 0 If we keep going, we will keep getting 0. So, reducing the number down to 0 looks like a base case. Every number will do this. So our base case becomes If the number is 0 return 0. The recursive case must take the rightmost digit and add it to the sum of the rest of the digits. To get the rightmost, again, look at the description. Here is an example 123 % 10 = 3 12 % 10 = 2 1 % 10 = 1 We take these two ideas and get the following pseudocode Int sumDigits(int num) { If(num is 0) { Return 0; } Return rightmost digit + sumDigits(rest of number) }

Perform a trace with num = 39312.