AP Computer Science A : Traversals
Study concepts, example questions & explanations for AP Computer Science A
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Example Questions
Example Question #1 : Computer Science
Consider the code below:
public static class BTNode {
public static final int PARSE_IN = 1;
public static final int PARSE_PRE = 2;
public static final int PARSE_POST = 3;
String name;
BTNode lPointer,rPointer;
public BTNode(String s) {
name = s;
lPointer = rPointer = null;
}
public void insert(String s) {
insert(this,s);
}
private static void insert(BTNode node,String s) {
int comparison = s.compareTo(node.name);
if(comparison < 0) {
if(node.lPointer != null) {
insert(node.lPointer,s);
} else {
node.lPointer = new BTNode(s);
}
} else if(comparison > 0) {
if(node.rPointer != null) {
insert(node.rPointer,s);
} else {
node.rPointer = new BTNode(s);
}
}
}
public ArrayList<String> parse(final int parseOrder) {
return parse(this,parseOrder);
}
private static ArrayList<String> parse(BTNode node, final int parseOrder) {
ArrayList<String> retVal = new ArrayList<String>();
if(node == null) {
return(retVal);
}
ArrayList<String> leftList = parse(node.lPointer,parseOrder);
ArrayList<String> rightList = parse(node.rPointer,parseOrder);
if(parseOrder == PARSE_PRE) {
retVal.add(node.name);
retVal.addAll(leftList);
retVal.addAll(rightList);
} else if (parseOrder == PARSE_POST) {
retVal.addAll(leftList);
retVal.addAll(rightList);
retVal.add(node.name);
} else {
retVal.addAll(leftList);
retVal.add(node.name);
retVal.addAll(rightList);
}
return retVal;
}
}
public static void main(String[] args) {
String[] names = {"Hervaeus","Peter Auriol","Guiral","Felix","Lila","Lola","Yippy","Yiiiipppy","Acton","Pierce","Betty"};
BTNode node = new BTNode(names[0]);
for(int i = 1; i < names.length; i++) {
node.insert(names[i]);
}
ArrayList<String> traversedNames = node.parse(BTNode.PARSE_IN);
for(String s : traversedNames) {
System.out.println(s);
}
}
What is the output for this method?
There is an error in the recursion in BTNode.
Acton
Betty
Felix
Guiral
Hervaeus
Lila
Lola
Peter Auriol
Pierce
Yiiiipppy
Yippy
Betty
Acton
Felix
Guiral
Lola
Lila
Pierce
Yiiiipppy
Yippy
Peter Auriol
Hervaeus
Hervaeus
Guiral
Felix
Acton
Betty
Peter Auriol
Lila
Lola
Yippy
Yiiiipppy
Pierce
Peter Auriol
Hervaeus
Guiral
Acton
Betty
Felix
Lila
Lola
Yippy
Pierce
Yiiiipppy
Acton
Betty
Felix
Guiral
Hervaeus
Lila
Lola
Peter Auriol
Pierce
Yiiiipppy
Yippy
The code given is a standard implementation of a binary tree containing an insert and a parse method. For now, you can merely pay attention to your parse logic, particularly just the logic that will be invoked for the indicator value PARSE_IN. (Notice that this is merely in the else of the parse method. This is a "catch all" / default in case a bad value is given for the parse order.)
retVal.addAll(leftList);
retVal.add(node.name);
retVal.addAll(rightList);
This is just the standard binary tree style of parsing based on our insert. The smaller items are on the left of the current node, and the larger ones are on the right of it. Thus, you have:
List of all smaller values + This current value + List of all larger values
Recursively, this will end with an ordered list, which is just what you need.
Example Question #2 : Computer Science
Consider the following code:
import java.util.ArrayList;
public class MethodClass5 {
public static class BTNode {
public static final int PARSE_IN = 1;
public static final int PARSE_PRE = 2;
public static final int PARSE_POST = 3;
String name;
BTNode lPointer,rPointer;
public BTNode(String s) {
name = s;
lPointer = rPointer = null;
}
public void insert(String s) {
insert(this,s);
}
private static void insert(BTNode node,String s) {
int comparison = s.compareTo(node.name);
if(comparison < 0) {
if(node.lPointer != null) {
insert(node.lPointer,s);
} else {
node.lPointer = new BTNode(s);
}
} else if(comparison > 0) {
if(node.rPointer != null) {
insert(node.rPointer,s);
} else {
node.rPointer = new BTNode(s);
}
}
}
public ArrayList<String> parse(final int parseOrder) {
return parse(this,parseOrder);
}
private static ArrayList<String> parse(BTNode node, final int parseOrder) {
ArrayList<String> retVal = new ArrayList<String>();
if(node == null) {
return(retVal);
}
ArrayList<String> leftList = parse(node.lPointer,parseOrder);
ArrayList<String> rightList = parse(node.rPointer,parseOrder);
if(parseOrder == PARSE_PRE) {
retVal.add(node.name);
retVal.addAll(leftList);
retVal.addAll(rightList);
} else if (parseOrder == PARSE_POST) {
retVal.addAll(leftList);
retVal.addAll(rightList);
retVal.add(node.name);
} else {
retVal.addAll(leftList);
retVal.add(node.name);
retVal.addAll(rightList);
}
return retVal;
}
}
public static void main(String[] args) {
String[] names = {"Thomas Aquinas","Thomas Cajetan","Thomas Prufer","Thomas the Tank Engine","Thomas the Bread-Eater"};
BTNode node = new BTNode(names[0]);
for(int i = 1; i < names.length; i++) {
node.insert(names[i]);
}
ArrayList<String> traversedNames = node.parse(BTNode.PARSE_POST);
for(String s : traversedNames) {
System.out.println(s);
}
}
}
What is the output for the main method above?
Thomas the Tank Engine
Thomas Prufer
Thomas the Bread-Eater
Thomas Aquinas
Thomas Cajetan
Thomas Aquinas
Thomas the Tank Engine
Thomas Prufer
Thomas the Bread-Eater
Thomas Cajetan
Thomas Aquinas
Thomas Cajetan
Thomas Prufer
Thomas the Bread-Eater
Thomas the Tank Engine
Thomas the Bread-Eater
Thomas the Tank Engine
Thomas Prufer
Thomas Cajetan
Thomas Aquinas
Thomas the Bread-Eater
Thomas Aquinas
Thomas the Tank Engine
Thomas Prufer
Thomas Cajetan
Thomas the Bread-Eater
Thomas the Tank Engine
Thomas Prufer
Thomas Cajetan
Thomas Aquinas
This code is a standard implementation of a Binary Tree class. The call that we are making in main is meant to parse the tree (traverse it) in post-fix order. This means that we will always look to the left of each node first, then to the right, then finally to the value we are at. However, the tree is unbalanced, so the parsing will do much more right traveral than anything else. See the following sequence of inserts:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Now, your traversal path will look like this:
Interestingly, this means that you will end up with a list that is in reverse order. This is only the case for the given data as it happened to have been inserted.
Example Question #1 : Computer Science
recur is defined as follows:public int recur(int x){if (x <= 1) {return 1;}else{return x + recur(x/2);}recur called in the following declaration?int num = recur(6);1
3
5
4
2
3
The first call is in the declaration. Because 6 > 1, it calls recur, which makes the total 2.
Next, it calls recur(6/2). Because 3 > 1, it calls recur again, which makes the total 3.
Next, it calls recur(3/2). Because it's dividing integers, this evaluates to recur(1).
Because 1 <= 1, it doesn't call recur anymore, so that total number of calls is 3.
Example Question #2 : Computer Science
Which of the following code performs a multiplication by 5 of the elements of the array defined as:
int[][] vals = new int[50][100];
Presume that the array has been properly initialized and filled with values.
for(int i = 0; i < vals.length;i++) {
vals[i][0...99] *= 5;
}
for(int i = 0; i < vals.length;i++) {
vals[i] *= 5;
}
for(int i = 0; i < vals.length;i++) {
for(int j = 0; j < vals[i].length; j++) {
vals[i][j] *= 5;
}
}
for(int i = 0; i < vals.length;i++) {
for(int j = 0; j < vals.length; j++) {
vals[i][j] *= 5;
}
}
for(int i = 0; i < vals.length;i++) {
for(int j = 0; j < vals[i].length; j++) {
vals[j][j] *= 5;
}
}
for(int i = 0; i < vals.length;i++) {
for(int j = 0; j < vals[i].length; j++) {
vals[i][j] *= 5;
}
}
What we are looking for in this problem is a standard traversal of a 2D array. When you do this, you need to make sure that you iterate through both the rows and the columns. In order to do this, you first must set up a loop like:
for(int i = 0; i < vals.length;i++) {...
The value of vals.length indicates the number of rows in the 2D array.
Now, for each row, you have a certain number of columns. (A 2D array is like an "array of arrays".) Thus, for each row, you need to run through the entire set of columns for that row:
for(int j = 0; j < vals[i].length; j++) { ...
Notice that none of the incorrect questions use vals[i].length in this way. Thus, none of them iterate through the two dimensions of the array correctly.
Example Question #1 : Traversals
TREE TRAVERSALS
Given the following tree structure:
What is the pre-order traversal of the tree?
1 2 4 5 3 6 8 9 5 7
5 4 2 1 3 6 7 8 9 5
1 2 3 4 5 5 6 7 8 9
1 2 3 4 6 7 5 8 9 5
1 2 4 5 3 6 8 9 5 7
When a preorder traversal is done, you go through the following:
1. ROOT
2. LEFT CHILD
3. RIGHT CHILD
Therefore, starting at the root (1), we go to the LEFT node (2). That node however, is the root/parent of other nodes, so we go left again (4). That node is a parent/root, so we go left (5). Now it's time to go to the right; however, only node 1 has a right child, so going to the right of one we land at (3). That node is the parent/root of more children so we go left (6). Node 6 doesn't have any left children, but it does have a right so we go right (8). Node 8 has a left child to traverse (9) and then a right (5). And we finish the traversal by going to the right of three (7). This is a rather complicated example; however, make sure you can traverse the tree properly.
In the above image, you can see the approach of the pre-order traversal. First you go through the roots, then the left trees, then the right trees. Make sure that if you're guiding yourself by the image above, that you don't repeat the nodes that you already traversed. Understanding the algorithm is key.
Example Question #11 : Operations On Data Structures
Traverse and print out this list
List<Integer> integers = new ArrayList<Integer>();
integers.add(1);
integers.add(2);
integers.add(3);
for (int i = 0; i < integers.length-1; i++) {
System.out.println(integers.get(i));
}
for (int i = 0; i < integers.size()-1; i++) {
System.out.println(integers.get(i));
}
System.out.println(integers.get(0));
System.out.println(integers.get(1));
System.out.println(integers.get(2));
for (int i = 0; i < integers.size()-1; i++) {
System.out.println(integers[i]);
}
for (int i = 0; i < integers.size()-1; i++) {
System.out.println(integers.get(i));
}
Use a for loop to traverse the list. The .size() method is used for Lists as opposed to the .length method for arrays. The .get() method is used for Lists as opposed to accessing the index for arrays.
Example Question #11 : Standard Operations & Algorithms
Suppose you are given an array of integers:
int array[] = {1,2,3,4,5};
and the following method:
public static void printArray(int[] arr)
{
for (int a = 0; r < arr.length-1; a++)
{
if (a%2==0)
{
System.out.println(arr[a]);
}
}
}
After the method cacll printArray(array) is called, the output would be:
2
4
1
2
3
4
5
1
3
1
3
5
2 4 6
1
3
The correct answer is:
1
3
The important line here is the if statement, which only executes if (a%2==0) is true, or if the loop counter divided by 2 is 0. A possible error is to divide each element of the array by 2 instead of the loop counter. So the println statement will only execute if the loop counter is even, which happens on the 1st itteration (a=0) and the 3rd itteration (a=2). The for loop ends at arr.length-1, which means that a takes on a maximum value of 3, and not 4 (choosing the option 1 3 5 would mean you did not notice the bounds on the for loop). Finally, println prints a new line every time, so, it is not possible for all the integers to be on one line, as println is executed twice.
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