This is my code so far, but idk why Im getting this error at the temp.getLeft() = null; part.
public Node removeLeaf(){
return removeLeaf(this.root);
}
public Node removeLeaf(Node node){
Node temp;
if(node.getLeft()==null && node.getRight()==null){
temp = node.parent;
}
if(temp == null){
return null;
}
if(temp.getLeft().equals(node)&& temp.getLeft()!=null){
temp.getLeft() = null;
}
}
Related
(source, destination) and it's type (tree, back, forward, cross)?
Here you go. Code in Java
import java.util.ArrayList;
import java.util.List;
class Node {
public String name;
public List<Node> connections = new ArrayList<>();
boolean visited = false;
Node(String name) {
this.name = name;
}
}
class DFS {
// Main part.
public static void search(Node root) {
if (root == null) {
return;
}
root.visited = true;
for (Node node : root.connections) {
if (!node.visited) {
// Print result.
System.out.println(root.name + "->" + node.name);
search(node);
}
}
}
}
public class App {
public static void main(String[] args) {
Node a = new Node("a");
Node b = new Node("b");
Node c = new Node("c");
Node d = new Node("d");
Node e = new Node("e");
a.connections.add(b);
b.connections.add(a);
b.connections.add(c);
b.connections.add(d);
c.connections.add(b);
c.connections.add(d);
d.connections.add(b);
d.connections.add(c);
d.connections.add(e);
DFS.search(d);
}
}
Nice question.
This is the solution based on the source you posted as comment.
IMPORTANT: There is an error on the start/end table, third row third column should be "end[u] < end[v]" instead of "end[u] > end[v]"
void main(G, s){
for each node v in G{
explored[v]=false
parent[v]=null
start[v]=end[v]=null
}
Global clock = 0
DFS(G, s, s)
}
void DFS(G, s, parent){
explored[s] = true;
parent[s] = parent
start[s]=clock
clock++
for each u=(s,v){
if(explored[v] == false){
DFS(G, v)
}
print (s + "-->" + v +"type: " + getType(s,v))
}
end[s]=clock
clock++
}
String getType(s, v){
if(start[s]<start[v]){
if(end[s]>end[v]) return "Tree edge"
else return "Forward edge"
else{
if(end[s]<end[v]) return "Back edge"
else return "Cross edge"
}
}
void Btree<T>::InsertNode2(T data, BtreeNode* root)
{
if (root==NULL)
{
root = new BtreeNode (data);
return ;
}
if (data <= root->data)
InsertNode2(data, root->leftchild);
else
InsertNode2(data, root->rightchild);
}
Why isn't it right? The root can't be assigned correctly. After calling the function it's still NULL.
It almost correct. Just, as said by #DragonMoon, all you need to do is to pass root by reference
void Btree<T>::InsertNode2(T data, BtreeNode &* root)
{
if (root==NULL)
{
root = new BtreeNode (data);
return ;
}
if (data <= root->data)
InsertNode2(data, root->leftchild);
else
InsertNode2(data, root->rightchild);
}
I was wondering if in a way to avoid having to deal with the root as a special case in a Binary Search Tree I could use some sort of sentinel root node?
public void insert(int value) {
if (root == null) {
root = new Node(value);
++size;
} else {
Node node = root;
while (true) {
if (value < node.value) {
if (node.left == null) {
node.left = new Node(value);
++size;
return;
} else {
node = node.left;
}
} else if (value > node.value) {
if (node.right == null) {
node.right = new Node(value);
++size;
return;
} else {
node = node.right;
}
} else return;
}
}
}
For instance, in the insert() operation I have to treat the root node in a special way. In the delete() operation the same will happen, in fact, it will be way worse.
I've thought a bit regarding the issue but I couldn't come with any good solution. Is it because it is simply not possible or am I missing something?
The null node itself is the sentinel, but instead of using null, you can use an instance of a Node with a special flag (or a special subclass), which is effectively the null node. A Nil node makes sense, as that is actually a valid tree: empty!
And by using recursion you can get rid of the extra checks and new Node littered all over (which is what I presume is really bothering you).
Something like this:
class Node {
private Value v;
private boolean is_nil;
private Node left;
private Node right;
public void insert(Value v) {
if (this.is_nil) {
this.left = new Node(); // Nil node
this.right = new Node(); // Nil node
this.v = v;
this.is_nil = false;
return;
}
if (v > this.v) {
this.right.insert(v);
} else {
this.left.insert(v);
}
}
}
class Tree {
private Node root;
public Tree() {
root = new Node(); // Nil Node.
}
public void insert(Value v) {
root.insert(v);
}
}
If you don't want to use recursion, your while(true) is kind of a code smell.
Say we keep it as null, we can perhaps refactor it as.
public void insert(Value v) {
prev = null;
current = this.root;
boolean left_child = false;
while (current != null) {
prev = current;
if (v > current.v) {
current = current.right;
left_child = false;
} else {
current = current.left;
left_child = true;
}
}
current = new Node(v);
if (prev == null) {
this.root = current;
return;
}
if (left_child) {
prev.left = current;
} else {
prev.right = current;
}
}
The root will always be a special case. The root is the entry point to the binary search tree.
Inserting a sentinel root node means that you will have a root node that is built at the same time as the tree. Furthermore, the sentinel as you mean it will just decrease the balance of the tree (the BST will always be at the right/left of its root node).
The only way that pops in my mind to not manage the root node as a special case during insert/delete is to add empty leaf nodes. In this way you never have an empty tree, but instead a tree with an empty node.
During insert() you just replace the empty leaf node with a non-empty node and two new empty leafs (left and right).
During delete(), as a last step (if such operation is implemented as in here) you just empty the node (it becomes an empty leaf) and trim its existing leafs.
Keep in mind that if you implement it this way you will have more space occupied by empty leaf nodes than by nodes with meaningful data. So, this implementation has sense only if space is not an issue.
The code would look something like this:
public class BST {
private Node root;
public BST(){
root = new Node();
}
public void insert(int elem){
root.insert(elem);
}
public void delete(int elem){
root.delete(elem);
}
}
public class Node{
private static final int EMPTY_VALUE = /* your empty value */;
private int element;
private Node parent;
private Node left;
private Node right;
public Node(){
this(EMPTY_VALUE, null, null, null);
}
public Node(int elem, Node p, Node l, Node r){
element = elem;
parent = p;
left = l;
right = r;
}
public void insert(int elem){
Node thisNode = this;
// this cycle goes on until an empty node is found
while(thisNode.element != EMPTY_VALUE){
// follow the correct path for the insertion here
}
// insert new element here
// thisNode is an empty node at this point
thisNode.element = elem;
thisNode.left = new Node();
thisNode.right = new Node();
thisNode.left.parent = thisNode;
thisNode.right.parent = thisNode;
}
public void delete(int elem){
// manage delete here
}
}
I am looking at the code to do this in CC150. One of its method is as follows, it does this by retrieving the tail of the left sub-tree.
public static BiNode convert(BiNode root) {
if (root == null) {
return null;
}
BiNode part1 = convert(root.node1);
BiNode part2 = convert(root.node2);
if (part1 != null) {
concat(getTail(part1), root);
}
if (part2 != null) {
concat(root, part2);
}
return part1 == null ? root : part1;
}
public static BiNode getTail(BiNode node) {
if (node == null) {
return null;
}
while (node.node2 != null) {
node = node.node2;
}
return node;
}
public static void concat(BiNode x, BiNode y) {
x.node2 = y;
y.node1 = x;
}
public class BiNode {
public BiNode node1;
public BiNode node2;
public int data;
public BiNode(int d) {
data = d;
}
}
What I don't understand is the Time Complexity the author gives in the book O(n^2). What I came up with is T(N) = 2*T(N/2) + O(N/2), O(N/2) is consumed by the getting tail reference because it needs to traverse a list length of O(N/2). So by Master Theorem, it should be O(NlogN). Did I do anything wrong? Thank you!
public static BiNode convert(BiNode root) {//worst case BST everything
if (root == null) { // on left branch (node1)
return null;
}
BiNode part1 = convert(root.node1);//Called n times
BiNode part2 = convert(root.node2);//Single call at beginning
if (part1 != null) {
concat(getTail(part1), root);// O(n) every recursive call
} // for worst case so 1 to n
// SEE BELOW
if (part2 != null) {
concat(root, part2);
}
return part1 == null ? root : part1;
}
public static BiNode getTail(BiNode node) {//O(n)
if (node == null) {
return null;
}
while (node.node2 != null) {
node = node.node2;
}
return node;
}
public static void concat(BiNode x, BiNode y) {//O(1)
x.node2 = y;
y.node1 = x;
}
SAMPLE TREE:
4
/
3
/
2
/
1
As you can see, in the worst case scenario (Big-Oh is not average case), the BST would be structured using only the node1 branch(es). Thus, the recursion would have a have to run getTail() with '1 + 2 + ... + N' problem sizes to complete the conversion.
Which is O(n^2)
I have run into a few problems when trying to get the camera to work accordingly... The camera Demo Works on the 8520 device (Has a memory Card) but does not work on the 9780 device (Has No Memory Card) the error given
ERROR Class java.lang.ArrayOutOfBoundsException :index 0>=0
My code Sample:
public class MyScreen extends MainScreen{
Player _p;
VideoControl _videoControl;
FileConnection fileconn;
String PATH;
String GetfileName;
LabelField GetPhotofileName = new LabelField("",LabelField.FOCUSABLE){
protected boolean navigationClick(int status, int time){
Dialog.alert("Clicked");
return true;
}
};
public static boolean SdcardAvailabulity() {
String root = null;
Enumeration e = FileSystemRegistry.listRoots();
while (e.hasMoreElements()) {
root = (String) e.nextElement();
if( root.equalsIgnoreCase("sdcard/") ) {
}else if( root.equalsIgnoreCase("store/") ) {
}
}
class MySDListener implements FileSystemListener {
public void rootChanged(int state, String rootName) {
if( state == ROOT_ADDED ) {
if( rootName.equalsIgnoreCase("sdcard/") ) {
}
} else if( state == ROOT_REMOVED ) {
}
}
}
return true;
}
protected boolean invokeAction(int action){
boolean handled = super.invokeAction(action);
if(SdcardAvailabulity()){
PATH = System.getProperty("fileconn.dir.memorycard.photos")+"Image_"+System.currentTimeMillis()+".jpg";//here "str" having the current Date and Time;
} else {
// PATH = System.getProperty("file:///store/home/user/pictures/")+"Image_"+System.currentTimeMillis()+".jpg";
PATH = System.getProperty("fileconn.dir.photos")+"Image_"+System.currentTimeMillis()+".jpg";
}
if(!handled){
if(action == ACTION_INVOKE){
try{
byte[] rawImage = _videoControl.getSnapshot(null);
System.out.println("----------1");
fileconn=(FileConnection)Connector.open(PATH);
System.out.println("----------2");
if(fileconn.exists()){
fileconn.delete();
System.out.println("----------3");
}
fileconn.create();
System.out.println("----------4");
OutputStream os=fileconn.openOutputStream();
System.out.println("----------5");
os.write(rawImage);
GetfileName =fileconn.getName();
System.out.println("----------6");
System.out.println("GetfileName----------"+GetfileName);
fileconn.close();
System.out.println("----------7");
os.close();
Status.show("Image is Captured",200);
GetPhotofileName.setText(GetfileName);
System.out.println("----------8");
if(_p!=null)
_p.close();
System.out.println("----------9");
}catch(Exception e){
if(_p!=null){
_p.close();
}
if(fileconn!=null){
try{
fileconn.close();
}catch (IOException e1){
//if the action is other than click the trackwheel(means go to the menu options) then we do nothing;
}
}
}
}
}
return handled;
}
public MyScreen(){
setTitle("Camera App");
try{
System.out.println("Debug------------10");
_p = javax.microedition.media.Manager.createPlayer("capture://video?encoding=jpeg&width=1024&height=768");
_p.realize();
_videoControl = (VideoControl) _p.getControl("VideoControl");
System.out.println("Debug------------11");
if (_videoControl != null){
Field videoField = (Field) _videoControl.initDisplayMode (VideoControl.USE_GUI_PRIMITIVE, "net.rim.device.api.ui.Field");
_videoControl.setDisplayFullScreen(true);
System.out.println("Debug------------12");
_videoControl.setVisible(true);
_p.start();
System.out.println("Debug------------13");
if(videoField != null){
add(videoField);
System.out.println("Debug------------14");
}
}
}catch(Exception e){
if(_p!=null) {
_p.close();
}
Dialog.alert(e.toString());
}
add(GetPhotofileName);
}
}
on the 8520 (Has a Memory Card) the code works fine on the 9780 (Has no Memory Card) the the code stops at "System.out.println("debug---1")", can anyone please tell me if you can see any problem with my code???
public static boolean SdcardAvailabulity() {
String root = null;
Enumeration e = FileSystemRegistry.listRoots();
while (e.hasMoreElements()) {
root = (String) e.nextElement();
if( root.equalsIgnoreCase("sdcard/") ) {
return true;
}else if( root.equalsIgnoreCase("store/") ) {
return false;
}
}
class MySDListener implements FileSystemListener {
public void rootChanged(int state, String rootName) {
if( state == ROOT_ADDED ) {
if( rootName.equalsIgnoreCase("sdcard/") ) {
}
} else if( state == ROOT_REMOVED ) {
}
}
}
return true;
}
This is the sollution, My "SD card availability" code only returned true which caused the picture not to save when the blackberry had no memory card inserted. # Eugen Martynov Please read through the code and you will see it is there :)