I am trying to change my recursive insert method of the BST into non-recursive( maybe While loop)
The reason for this changing because I want to see if it is possible.
Here is the code of insertion:
public void insert(String value)
{
//The node is stored in the root
root = insert(value,root);
}
private Character insert(String value,Character current)
{
if(current == null)
{
//Add the root if the tree empty
current = new Character(value);
}
else
//If the value that we want to insert < root value, then keep going to left till
//it's empty then inserted at left end. Done by recursion
if(value.compareTo(current.getElement())<=-1)
{
current.setLeft(insert(value,current.getLeft()));
}
else
//If the value that we want to insert > root value, then keep going to right till
//it's empty then inserted at right end. Done by recursion
if(value.compareTo(current.getElement())>=1)
{
current.setRight(insert(value,current.getRight()));
}
else
{
//Else, the number we want to insert in already in the tree
System.out.println("Duplicate numbers inserted" + current.getElement());
}
//returning the node tree so that we store it in the root
return current;
}
Could I change this code into non recursive ?
Cheers
Yes, but you need to alter the data structure a little bit to make it works.
The node has to know its left child and right child.
The algorithm looks like this:
current = root;
while(current != null){
if(value.compareTo(current.getElement())<=-1)
{
current = current.left_child;
}else if(value.compareTo(current.getElement())>=1){
current = current.right_child;
}else{
// Duplication
}
}
Actually there are some good examples before, you may want to check those out first:
Write a non-recursive traversal of a Binary Search Tree using constant space and O(n) run time
Nonrecursive/Iterative Binary Search Tree in C (Homework)
Yes, you could define your insert function non-recursively.
However, to do this, your insert function will have to define in-order traversal iterator for BST, which is recursively defined.
I believe there is a way to define in-order traversal non-recursively, but depending on implementation this can be very inefficient.
BST itself is basically recursively defined, and it is always efficient to define your insert function recursively. (I could write some pseudo-code if you really need it, but I think it is kind of meaningless and I do not know about the implementation detail of your in-order traversal iterator)
Please don't forget to select this as an answer :-)
Insert using while loop
public Node insert(Node root,int n) {
while (true) {
if (root.data>n) {
if (root.left==null) {
root.left= new Node(n);
return (root.left);
}
root=root.left;
}
else if (root.data<n) {
if (root.right == null) {
root.right= new Node(n);
}
}
}
}
Related
I have the following code to insert into the binary tree:
public void insert(T item) {
root = insert(item, root);
}
private Node insert(T item, Node node) {
if(node == null){
return new Node(item, null, null);
} else {
if(item.compareTo(node.item) > 0) {
node.rightChild = insert(item, node.rightChild);
} else {
node.leftChild = insert(item, node.leftChild);
}
}
return node;
}
the code works fine, I have tested it
my question is, how come the root is never changed since in the public function I assigned the returned node from the private function to the root
Thank you!
The public insert is just an interface into the recursive method which rebuilds the tree as the stack unwinds ending where you started, at the root (root = ...). Except for the first insert, you go left or right until you insert at the leaf level. Without any balancing, you will have the same root (assuming no removals) for the lifetime of the tree. Therefore, the only time the root changes on insert is when it's empty.
Note: There's also a matter of what happens when a node is inserted with an existing value; do you discard it, allow duplicates, or swap the objects? That's an implementation detail.
I read your code again and your code is correct. The return value is always the root element because your frist function call is insert(item, root) and the return value is what you give.
class Link{
private int value;
private Link next;
}
I am asked to write a recursive method to delete last occurrence of a certain value, say 4.
before 2->3->4->5->4->2
after 2->3->4->5->2
The last occurrence only. I know how to delete all occurrence but I can't tell if its the last occurrence. No helper method is allowed.
The one to delete all occurrence
public Link deleteAll(){
if (next == null){
return value==4？ null:this;
}else{
if (value == 4){
return next.deleteAll();
}
next = next.deleteAll();
return this;
}
}
You can declare a pointer to the last occurred node and delete that node when reached the last element in list. Following steps explains that -
Declare two pointers one is next as in your above code another can be temp.
Iterate through list using next like you doing in deleteAll method above.
If you find the node you looking for assign that node to temp.In your case 4.
When next is null you reached the end of list now delete, whatever node is in temp delete that node. If temp is still null than no node found in given key.
EDIT:
Possible pseudo Code in case of recursion:
public void deleteLast(Node node,Node temp,Node prev, int data)
{
if(node==null)
{
if(temp!=null && temp.next.next!=null){
temp.next = temp.next.next;}
if(temp.next.next==null)
temp.next = null;
return;
}
if(node.data==data)
{
temp = prev;
}
prev = node;
deleteLast(node.next, temp, prev, int data);
}
Above code should be able to solve your problem. I made some edit in my approach which should be obvious from the code but let me describe it below
I added a prev pointer. Because if we want to delete a particular node we need to assign its next to prev node's next.So, we need the prev node not the node that we want to delete.
I think this change will follow in iterative approach too.
Not really answering your exact question, but as an alternative option, you might consider the following.
Write a recursive method to delete the first occurrence of a specified value, something like this:
public Link deleteFirst(int target) {
if (value == target) {
return next;
}
next = (next == null) ? null : next.deleteFirst(target);
return this;
}
Then you could write a reverse() method as either an iterative or recursive method as you see fit. I haven't included this, but googling should show some useful ideas.
Finally the method to remove the last occurrence of a value from the linked list could then be written like this:
public Link deleteLast(int target) {
return reverse().deleteFirst(target).reverse();
}
Note that as long as your reverse() method is linear complexity, this operation will be linear complexity as well, although constants will be higher than necessary.
The trick is to do the work on the way back -- there is no need for additional parameters, helpers or assumptions at all:
Link deleteLast(int target) {
if (next == null) {
return null;
}
Link deleted = next.deleteLast(target);
if (deleted == null) {
return value == target ? this : null;
}
if (deleted == next) {
next = deleted.next;
}
return deleted;
}
I am trying to write a code to convert a binary tree to a lists of nodes with same depth. If a tree has depth d, then d lists will be created. The logic is to do in-order traversal and add the currently traversed node to the list of appropriate depth.
public void treeToListofNodesByLevel(Node<T> n,int depth, ArrayList<LinkedList<Node<T>>> treeList){
if(n.right != null){
inOrderWithHeight(n.right, depth + 1);
}
if(treeList.size() >= depth){
treeList.add(depth, new LinkedList<Node<T>>() );
}
treeList.get(depth).add(n);
if(n.left != null){
inOrderWithHeight(n.left, depth + 1);
}
}
and then calling:
ArrayList<LinkedList<Node<T>>> result = new ArrayList<LinkedList<Node<T>>>();
treeToListofNodesByLevel(root, 0, result);
Will this work ? Are there any corner cases I am not handling ?
Also, right now I am passing the List of List to be returned by the method because I can not think of a way to initialize it in the method and returning it at then end while also maintaining the recursive structure. Is there a better way to do this ?
You have the general concept pretty much perfect. It will work, and should handle all cases.
However, you have a few errors in the details:
Your check for when to add a new list has the comparison in the wrong direction. It should be if (treeList.size() <= depth).
Each call to inOrderWithHeight() (which you haven't provided any code of) should be a recursive call to treeToListofNodesByLevel(). Keep the first two arguments as they are, and just pass the treeList for the third.
This one's more a style issue, but parameter types should generally be declared as the highest level type that satisfies what you actually need. There is no need here to specify ArrayList or LinkedList, any List will do. Change the treeList parameter's type to List<List<Node<T>>>.
For the matter of initializing the List inside the method while also using recursion, that's the sort of thing that implementation helper methods are for. Take the current body of treeToListofNodesByLevel and move it into a private method (with the recursive calls changed so the private method calls itself), let's call it treeToListofNodesByLevelHelper. Then change the current public method to this:
public List<List<Node<T>>> treeToListofNodesByLevel(Node<T> node) {
List<List<Node<T>>> result = new ArrayList<>();
treeToListofNodesByLevelHelper(node, 0, result);
return result;
}
I cannot understand what the method "inOrderWithHeight" is doing. What I do for this question (not optimized) is to traverse like BFS using two queues and for each iteration adding the nodes of that depth to the list of that iteration (each iteration is traversing one depth of the tree). Here is my code to do that for a binary tree as you supposed in your answer:
Queue<Node<T>> queue1 = new LinkedList<Node<T>>() ;
Queue<Node<T>> queue2 = new LinkedList<Node<T>>() ;
public void treeToListofNodesByLevel(Node<T> root, ArrayList<LinkedList<Node<T>>> treeList) {
if (root == null)
return;
int curr_depth = 0;
queue1.add(root);
while(!queue1.isEmpty()){
treeList.add(curr_depth, new LinkedList<Node<T>>());
while(!queue1.isEmpty()){
Node<T> node = queue1.remove();
treeList.get(curr_depth).add(node);
if(node.left != null) queue2.add(node.left);
if(node.right != null) queue2.add(node.right);
}
curr_depth++;
while(!queue2.isEmpty()){
Node<T> node = queue2.remove();
queue1.add(node);
}
}
}
I wrote this on the fly without syntax checking and may have compiler errors. But hopefully you get the idea.
Why you just use the same function the code will be more beautifull :
public void treeToListofNodesByLevel(BinaryTree node, int currentDepth, List<List<BinaryTree>> result){
// Check if the list for the currentDepth is created
if(result.get(currentDepth) != null){
result.add(currentDepth, new ArrayList<BinaryTree>())
}
// Add the current node to the list
result.get(currentDepth).add(node);
// Recursive the left node
if(node.left != null){
treeToListofNodesByLevel(node.left, currentDepth+1, result);
}
// Recursive the right node
if(node.right != null){
treeToListofNodesByLevel(node.right, currentDepth+1, result);
}
}
In your main function :
List<List<BinaryTree>> result = new ArrayList<>();
BinaryTree root = new BinaryTree();
treeToListofNodesByLevel(root, 0, result);
So I just wrote code for insertion of nodes in binary tree (NOT BST).
I noticed that every time the recursive insert returns a 'node', it is assigned to the initial node.
Does this mean, that the memory reference of root of this tree would change on the completion of each insert?
public void add(int data)
{
root=add(root,data);
}
public static BinaryNode add(BinaryNode node, int data) {
if(node==null)
{
node=new BinaryNode(data);
}
else {
///IF not 1st element, flow enters this part
if(node.left==null && node.right==null)
{
node.left=add(node.right,data);
}
else if (node.right == null) {
node.right=add(node.right, data);
} else {
node.left=add(node.left, data);
}
}
return node;
}
Within add the only time you change node is if the tree at that point is empty, so the answer is no except for the first insert.
However, note that you add a new level to the tree only on the left (first if condition) so the "tree" you build is highly unbalanced to the left. This isn't really a "tree", it's more like a strange linked list. Also, since you don't maintain any particular sequence it can't be better than a simple unordered list for searches.
I am currently taking a data structure class in college and we are learning about binary search trees using linked lists. We have gone over the inOrder method recursively but I wanted to try to do the method iteratively. After some research I realized I have to use a stack as i traverse through the tree. I am able to get the the end of the left most side of the tree, however traversing back up the tree is where I am having trouble. I have tried various version of my code but I keep ending up with with a nil pointer exception or it prints out of order.
public void inOrder(){
// implement this method using non-recursive solution
if(m_root==null){
return;
}
Stack<BSTNode> myStack= new Stack<BSTNode>();
BSTNode current=m_root;
while (current!= null){
myStack.push(current);
current=current.getLeft();
}
while (current!=null&& !myStack.isEmpty()){
current=myStack.peek();
System.out.print(current.getInfo()+" ");
myStack.pop();
if(current.getRight()!=null){
myStack.push(current);
}
}
}
From what I can see there are a few problems with your code in the second while loop. The idea you have is in the right direction however there some logic errors. The conditions you have are correct but should not be together, they should be separated. the code below achieve what you are looking for.
public void inOrder(){
// implement this method using non-recursive solution
if(m_root==null){
return;
}
Stack<BSTNode> myStack= new Stack<BSTNode>();
BSTNode current=m_root;
while (current!= null){
myStack.push(current);
current=current.getLeft();
}
while (!myStack.isEmpty()){
current=(BSTNode)myStack.pop();
System.out.print(current.getInfo()+" ");
if(current.getRight() != null){
current=current.getRight();
while (current!= null){
myStack.push(current);
current=current.getLeft();
}
}
}
}
For starters, you have in your code two major flaws: you leave the first while loop with current pointing to null and thus you never enter the second while loop. Moreover, I think this
if(current.getRight()!=null){
myStack.push(current);
}
should be corrected with myStack.push(current.getRight());, otherwise you're trying to push over and over just the element you popped and you would enter an endless loop. But even so the logic of the exploration is wrong since you'll never go to the left of a node you arrived from a right link.
I tried to start from your code and create a working inorder traversal:
public void inOrder(){
Stack<BSTNode> myStack= new Stack<BSTNode>();
Set<BSTNode> visited = new HashSet<BSTNode>();
BSTNode current = m_root;
if(current!= null)
myStack.push(current);
while (!myStack.isEmpty()){
current = myStack.peek();
if(current.getLeft()!= null && !visited.contains(current.getLeft()))
myStack.push(current.getLeft());
else{
//here you explore the parent node
System.out.print(current.getInfo()+" ");
visited.add(current);
myStack.pop();
//and then you see if it has children on the right
if(current.getRight()!=null && !visited.contains(current.getRight))
myStack.push(current.getRight());
}
}
}