Develop Reference

delete node after specified node in doubly linked list. (recursively)

I have this recursive function that is supposed to delete the node that comes after the specified one in
a doubly linked list. However My method Isn't deleting anything. I am having trouble with rearranging the values in the list. Any ideas?
private void deleteAfterThis(T data, Node headAux) {
if(headAux == null) {
return;
}
Node deleteAfter = new Node(data);
Node target = deleteAfter.next;
if(target == null) {
return;
}
if(deleteAfter.prev == null){
if(target != tail && target==headAux) {
deleteAfter.next = target.next;
target.next.prev = deleteAfter;
size--;
deleteAfterThis(data, headAux.next);
}
else if(target == tail && target == headAux) {
deleteAfter.next = null;
deleteAfter = tail;
size--;
return;
}
}
else if(deleteAfter.prev != null) {
if(target != tail && target == headAux) {
deleteAfter.next = target.next;
target.next.prev = deleteAfter;
size--;
deleteAfterThis(data, headAux.next);
}
else if( target == tail && target == headAux) {
deleteAfter.next = null;
deleteAfter = tail;
size--;
return;
}
}
deleteAfterThis(data, headAux.next);
}

One mistake I see right off the bat is that you should not be creating a completely new node for deleteAfter. Intuitively, does it make sense to have to create a new node when attempting to delete one? I'll assume that, even knowing what the constructor for Node actually looks like, it sets the next and prev pointers to nodes to null. As a result, you'll keep recursively updating headAux until it's null without ever deleting anything. It seems what you want deleteAfter to be is headAux.next.
Another bug I see is that you've copy and pasted your checking logic twice - I recommend stepping through both cases and verifying if the logic should be identical within each block of the if and else-if blocks (it probably shouldn't).
Stepping into the logic, you should realize that the prev node of the current (headAux in your code) would be null only if headAux is the head of the list. Thus, it would be a bit more clear to rewrite the headAux.prev check as verifying if headAux is equal to the head of the linked list.
Looking at the actual deletion logic, it seems to make sense in the general case to me (assuming the fact that deleteAfter is the next node of headAux as stated above). You're making the prev of the node to be deleted to point to the deleted node's prev node and the next of the previous node (after having set the pointer, which I'm not too big of a fan of but it works) point to deleteAfter.
Lastly, when you do actually locate the node you'd like to delete, you probably shouldn't be calling the recursive function again. You already handle the setting of pointers correctly, so there shouldn't be a need to do so.
I would highly recommend you (re-)draw a sample use-case of your circular linked list and perform deletion on a sheet of paper before jumping to coding the edgecases (which aren't all handled here). The edge cases you should probably be aware of are the following: empty list, deleting head, deleting tail, single-node list. In your code, it seems that you did try to handle the deletion of the head (you'd have to remember to set the head afterwards). After getting that to work, making deletion work on other cases should be a breeze.

How does recursion traversing through trees

java version "1.8.0_92"
I am studying trees and how to traverse them using recursion. But I am confused about it works.
public void preOrder(BinaryTree root) {
if(root != null) {
System.out.println(root);
preOrder(root.leftChild); <-- this gets called and will start from the top of the function
preOrder(root.rightChild); <-- how can this get called if the top one will always calls itself?
}
}
I think that the second preOrder will never get called as the call above will always call itself so the second one will never be executed.

It does not always call itself. It keeps going until it bottoms out with a null leftChild. Then execution returns without doing anything -- it backs up one level and recurs on the rightChild of the lowest-level parent node. When that call also runs out of children, the execution returns from handling this lowest-level parent node, again backing up one call, and does the rightChild of that node ... until the entire tree has been traversed.

Think of it has as opening up a whole bunch of doors until you find what you're looking for, then you have to go back and close/check the other ones.
public void preOrder(BinaryTree root) {
if(root != null) {
System.out.println(root);
preOrder(root.leftChild); <-- this gets called and will start from the top of the function
preOrder(root.rightChild); <-- how can this get called if the top one will always calls itself?
}
}
We can replace the preOrder calls with the matching code to look like:
public void preOrder(BinaryTree root) {
if(root != null) { <-- note the if check
System.out.println(root);
if(root.leftChild!= null) { <-- note the if check
System.out.println(root.leftChild.leftChild);
preOrder(root.leftChild.leftChild); <- this expands into another if block
preOrder(root.leftChild.rightChild); <- this also expands
}
if(root.rightChild!= null) { <-- note the if check
System.out.println(root.rightChild.leftChild);
preOrder(root.rightChild.leftChild);
preOrder(root.rightChild.rightChild);
}
}
}
It keeps expanding outwards... until you hit the special "base" if conditional that stops the recursion. In this case it is when you've found a leaf node of the tree, i.e. node == null, it stops the expansion because the if conditional is no longer true and everything begins to collapse in on itself, or in other words it can continue executing normally down the block of code.

Recursion may seem to be confused when you just start to learn. Firstly, you can think of a subproblem. For preOder, it always prints out the root node, and left node, then right node. All you need is to solve a subproblem. (a basic simple tree below)
root
/ \
left right
/ \
... ...
Now go back to look at the code:
if(root != null) {
System.out.println(root);
preOrder(root.leftChild); // the left node now becomes another root, and keep making the next left node root until the next left node is null.
preOrder(root.rightChild); // this line of code won't be executed until the last preOrder function call its root == null(hit the condition.)
}
Trust recursion; it will always do the rest of part for you if your condition is correct. I agree to run debug mode to have a better understanding. It is very useful to learn how to use "step in" when you try to understand how code works.

Once root.leftChild becomes null (ie, when you get to a leaf of the tree), preOrder will be called like so: preOrder(null). When this happens, the condition will evaluate to false, and recursion unwind and stop, at which point preOrder(root.rightChild) will be evaluated.
Here's a call trace (in Scala):
case class BinaryTree(nodeName: String, leftChild: Option[BinaryTree], rightChild: Option[BinaryTree]) {
def preOrder(root: Option[BinaryTree], depth: Int = 0) {
root match {
case Some(root) => {
println(" " * depth + root.nodeName)
preOrder(root.leftChild, depth+4)
preOrder(root.rightChild, depth+4)
}
case None => println(" " * depth + "leaf")
}
}
}
val tree = new BinaryTree(
"root",
Some(new BinaryTree(
"a",
Some(new BinaryTree("aa", None, None)),
Some(new BinaryTree("ab", None, None)))),
Some(new BinaryTree(
"b",
Some(new BinaryTree("ba", None, None)),
Some(new BinaryTree("bb", None, None)))))
tree.preOrder(Some(tree))
root
a
aa
leaf
leaf
ab
leaf
leaf
b
ba
leaf
leaf
bb
leaf
leaf

Accessing External Nodes Of Binary Search Tree

In a "binary tree", an external node is a node that does not have any children, neither left or right, -correct me if I'm wrong-, and in a "binary search tree", an external nodes are always null, because according to my lecture notes, an internal nodes always have 2 children even if not created yet we assume children of that internal node are null.
So how can I access an external node if it's null?
I wrote this code as part of BST Node class :
/*
* Checks if this node is an internal node.
* Returns true if it is internal node, false otherwise.
*/
protected boolean isInternal(){
// TODO Put your code here
if(this!=null)
return true;
else
return false;
}
/*
* Checks if this node is an external node.
* Returns true if it is external node, false otherwise.
*/
protected boolean isExternal(){
if(this==null && this.left==null && this.right==null)
return true;
else return false;
}
Last method give me nullPointerException

In a "binary tree", an external node is a node that does not have any children, neither left or right
No. "External" here is used as an adjective. It means a node that is not part of the tree yet.
external nodes are always null
Er, a node can't "be" null. A node pointer can be null, but that just means it's not pointing to a node.
Nodes can have null pointers instead of pointers to child nodes. When it has no children (both pointers null), it's called a Leaf node.

Java: How to create a queue based on a Node class

I'm trying to create a queue using two classes, a Node class and a Queue class for an assignment. Here's the node class:
class Node
{
protected Node next;
public Node()
{
next = null;
}
}
This class basically links the data together using a Node.next object. I've successfully been able to create a stack with push() and pop(), because the two operations happen on the same end, so the point are just manipulated between pointing to a new added node, and the previous node.
However, I'm having some difficulties understanding the logic to create a queue based on a similar structure. My queue class looks something like this:
class Queue
{
private Node footer;
private Node header;
public Queue()
{
footer = null;
header = null;
}
public void add(Node newNode)
{
//Adds onto the queue from the 'footer' end.
}
public Node remove()
{
//Removes from the queue from the 'header' end.
}
Here's what I understand: (1)The header and the footer point to the same first node. (2) Subsequent adding should change the footer to point to the added nodes, but the header stays on the first node added. (3) The header should point to the next oldest node upon removal.
Here's what I can't figure out (and where it's different than popping from a stack). How do I get the header to point to the 'next oldest node', given that I have more than 2 nodes in this queue? I know I can do this if I link header.next to the next node in the queue, but how can I access the next node so that it can point to it?
I thought about how in add(), the newNode.next should point to the next newNode (reverse direction of a Stack), but this can't work because the next newNode isn't in existence yet.. Another idea was to modify the Node class to have a Node.previous for a way to point backwards, but I would be breaking specification for this assignment.
My instructor hinted something about "header.next will point for second item as header and footer point to first node initially," and that the way to do this is pretty simple. However, I've been drawing how this works, and I'm confused how the initial pointing to the same node will allow header.next to "automatically" point to the next oldest node, especially if more and more nodes are added and the footer eventually is separated from the header by more than 2 nodes. Is there something about OOP I'm not seeing?
Any help would be great!

To expand on, and offer a subtle alternative to #Sanjeev's answer (one that I think your instructor was hinting to):
Rather than using footer to store "actual" nodes, use it as a placeholder: Declare it as a final variable, initialize it in your constructor and make sure that either a) it's next node is always your header (this would be called a circular list), or its next node is null.
Can you see how this solves your "this can't work because the next newNode isn't in existence yet" problem: Of course you can't point the last node added to the next one that will be added before adding it - instead, you point it to this "dummy" node - which is a placeholder for the next node that will be added, when and if it is.
add(Node newestNode){
identify the last node added as the one whose next property is the footer.
change the next property of that node from footer to this new newestNode
set the next property of this new newestNode to footer
}
It would be preferable to identify that last node added as the one that footer is pointing to (rather than the one pointing to footer), which would be easy if you were allowed to have previous as well as next properties on nodes, but it sounds like you're not allowed to do that. Of course, since we're using footer as a "dummy node", we could simply use footer.next the way we would footer.previous and have it point backwards instead of forwards, but I'll leave you to consider how clean that would be. There are other options here that I'll leave you to consider as well.
How do I get the header to point to the 'next oldest node'`
The "oldest" node was the first one added. The "newest" node is the last one added. How is the order of the rest of the nodes stored? The same way it was in your Stack - by traversing a chain of references stored as instance variables on your nodes. The main point I want to make is that Stacks and Queues, when implemented as linked data structures, are much more similar than you seem to be thinking, at least from a : Iterating through any linked data structure is done by following traversing these links - don't get too hung up on the fact that you're "moving" in a different direction - the same basic principles apply:
Node remove(){
identify the "oldest" node as header.next.
Store a reference to that node so you can return it.
identify the "second oldest node" as header.next.next
change header.next to header.next.next
return the reference to the old header.next you saved above.
(Note that using header/footer as placeholders, rather than storing "actual" nodes in them as #Sanjeev suggests, is not necessary, but it'll make your life easier - for instance, by helping you avoid a lot of null checking)

Here is the sudo code that will help you get started.
public void add(Node newNode)
{
if footer is null ?
then
header = newNode and footer = newNode;
else
footer.next = newNode and footer = newNode;
end if
}
public Node remove()
{
Node returnMe = header;
if header is not null?
then
header = header.next
if header is null
then
footer = null;
endif
end if
return returnMe;
}

How do I get the header to point to the 'next oldest node', given that
I have more than 2 nodes in this queue? I know I can do this if I link
header.next to the next node in the queue, but how can I access the
next node so that it can point to it?
To make header point to that node, you only need do header = header.next. The reason is that Java objectt assignment is by reference. Since header.next is type of Node, header is type of Node, it will copy the address of header.next to header, i.e., header is advanced one place.
I thought about how in add(), the newNode.next should point to the
next newNode (reverse direction of a Stack), but this can't work
because the next newNode isn't in existence yet..
I think it is no need to considering reverse direction. The reason is because for adding , it is to add element to the tail/footer of the queue. The only special case is that the queue didn't have any elements (footer == header == null), 1 element : (footer = header = element), other case: header won't change, but you need to append element to footer, and then make footer point to the new node.
When only 1 element, footer.next == header.next == null

The first thing that you need to do is make sure the first node you create is the oldest so it should be the first to be removed from the Queue based on First In First Out (FIFO) principle to archive this you might need to modify you're add method to something like this, by the way this example is based on single linked list implementation.
void add(char new_data)
{
/* 1. alloc the Node and put data*/
Node new_Node = new Node(new_data);
/* 2. Make next of new Node as head */
new_Node.next = head;
/* 3. Move the head to point to new Node */
head = new_Node;
}
then you will need a remove method which will remove the oldest node on the list first remember in Queue the order of remove is First In First Out (FIFO)
that being said this remove method should help you
void remove()
{
// Store head node
Node temp = head, prev = null;
// If head node itself holds the key to be deleted
if (temp != null )
{
head = temp.next; // Changed head
return;
}
// Search for the key to be deleted, keep track of the
// previous node as we need to change temp.next
while (temp != null)
{
prev = temp;
temp = temp.next;
}
// If key was not present in linked list
if (temp == null) return;
// Unlink the node from linked list
prev.next = temp.next;
}
This worked for me on my linked list

Threaded Binary tree implementation from binary tree

I am working on assignment for school. It manly consists of a method that takes as input a binary tree and returns a double threaded tree. Eg(if left child = null then left child will be connected with preceding inorder parent and if right child = null the it will link to its inorder succesor. Now I have an idea for the implementation...
I iterate recursively trough the original BINARY tree and store into an array the inorder traversal. Now, because my teachers implementation requires that threaded trees be a different class from binary. I must traverse again trough the binary tree and convert each node from binaryNode to threadedNode thus having at the end a "duplicate" of the initial BinaryTree but as Threadedtree type. After I do this I traverse again trough this threadedTree and whenever i see a null left or right child I refer to the inorder arraylist and find the threads.
Now as you might have noticed this is extremely inefficient, i am essentially traversing the tree 3 times. My professor has stated that this could be done recursively with only one traversal, essentially converting to threadedNode and finding the threads all at once. I have tried multiple ways but i can not find one that works. Does anyone have any kind of tip or some way i can implement it? Thanks
This is the method as specified by the instructor
public static <T> ThreadedNode<T> thread(BinaryNode<T> root)
{
//threads a binary tree
}

The instructor is correct. One traversal is sufficient.
Traverse the original binary tree, creating new ThreadedNodes as you walk this tree.
public static <T> ThreadedNode<T> thread(BinaryNode<T> root) {
// We'll be keeping track of the "previous" node as we go, so use
// a recursive helper method. At first, there is no previous.
return threadHelper(root, null);
}
private static <T> ThreadedNode<T> threadHelper(BinaryNode<T> n, ThreadedNode<T> previous) {
// Create a new threaded node from the current root. Note that the threaded nodes
// are actually created in "preorder". Assume the ThreadedNode constructor sets
// the left, right, threadLeft, and threadRight fields to null.
ThreadedNode<T> t = new ThreadedNode<T>(n.getData());
// First go down the left side, if necessary.
if (n.getLeft() != null) {
// If there is a left child we have to descend. Note that as we go down the
// left side the previous doesn't change, until we start "backing up".
t.left = threadHelper(n.getLeft(), previous);
previous = t.left;
} else {
// If there is no left child, connect our left thread to the previous.
t.threadLeft = previous;
}
// Now before we go down the right side, see if the previous
// node (it will be in the left subtree) needs to point here.
if (previous != null && previous.right == null) {
previous.threadRight = t;
}
if (n.getRight() != null) {
// If there is a right child we can descend the right. As we go down we
// update previous to the current node. We do this just by passing the current
// node as the second parameter.
t.right = threadHelper(n.getRight(), t);
} else {
// No right child, no worries. We'll hook up our thread-right pointer
// later.
}
return t;
}
Consider the tree (A (B (D) ()) C). The first node you hit in an inorder traversal is D. There is no previous node. So save D as previous. Then the next node you hit is B. The previous node was D, which had no right child, so add a threaded right pointer from D to B. Then set previous to B and continue. Next you hit A. B had no right child, so add a threaded right link from B to A. A has a right child so continue, setting previous to A. The next node is C. C has no left child, so add a threaded left link from C to the current value of previous, which is A.

You could skip the second trip of traversal that you mention in your method. You could convert the nodes from BinaryNode to ThreadedNode on the fly. You'd still need to traverse twice, I think, for the inorder traversal, and for finding the threads and converting it to aThreadedTree.
For conversion on the fly, you could use the method that your instructor has given.
HTH!