I'm trying to populate a binary search tree using an insert(), however when every I 'print' the contents of my BST, I only get the last item that was inserted into the BST. What do I need to fix to make sure all of my value are retaining in the BST?
From debuggin, I think that my problem is that my public void insert() is setting the new value to root evertime it is called. I don't know how to fix it?
Here is my BST CLass:
public class BinarySearchTree<T extends Comparable<T>> {
private class BinarySearchTreeNode<E>{
public BinarySearchTreeNode<E> left, right;
private E data;
private BinarySearchTreeNode (E data) {
this.data = data;
}
}
private BinarySearchTreeNode<T> root;
public boolean isEmpty() {
return root == null;
}
private BinarySearchTreeNode<T> insert(T value, BinarySearchTreeNode<T> ptr) {
if (ptr == null){
ptr = new BinarySearchTreeNode<>(value);
return ptr;
}
int compare = value.compareTo(ptr.data); //when ptr != null, this line and below should execute for each bstStrings.inster(x)
/* pass the value (s1...sN) when compared to (ptr.data) to compare
* when value and ptr.data == 0 re
*/
if (compare == 0) {
return ptr;
}
if (compare < 0) {
while (ptr.left != null){
ptr = ptr.left;
if (ptr.left == null) {//found insertion point
BinarySearchTreeNode<T> node = new BinarySearchTreeNode<>(value);
ptr = ptr.left;
ptr = node;
return ptr;
}
}
}
else {
return insert(value, ptr.left);
}
if (compare > 0) {
if (ptr.right == null) {
BinarySearchTreeNode<T> node = new BinarySearchTreeNode<>(value);
ptr = ptr.right;
ptr = node;
return ptr;
}
}
else {
return insert(value, ptr.right);
}
return ptr;
}
public void insert(T value) {
root = insert(value, root); //****Where I believe the problem is******
}
private void printTree(BinarySearchTreeNode<T>node){
if(node != null){
printTree(node.left);
System.out.println(" " + node.data);
printTree(node.right);
}
}
public void printTree(){
printTree(root);
System.out.println();
}
}
For added context here is my Main() where I am calling the insert() and trying to insert strings into the BST:
public class Main {
public static void main(String[] args) {
BinarySearchTree<String> bstStrings = new BinarySearchTree<String>();
String s = "Hello";
String s1 = "World";
String s2 = "This Morning";
String s3 = "It's";
bstStrings.insert(s);
bstStrings.insert(s1);
bstStrings.insert(s2);
bstStrings.insert(s3); //the only inserted value that is printed below
bstStrings.printTree();
System.out.println();
System.out.println("You should have values above this line!");
}
}
Lastly my console output:
It's
You should have values above this line!
Some hints:
I don't see any recursive calls inside insert. How would you traverse the BST without appropriate recursive calls (into the left or right subtree of the current node based on the value)? I do see some commented out code that looks like it would perform those calls. Why are they commented out?
You're returning the newly inserted node, which you're then setting as root. This will set the root to point to the new node every time. I don't think that's what you want.
If you're trying to handle the special case where the tree is empty, all you need to do is check to see if root is null, then set the new node to that.
There really is no need to return ptr. Since your BST maintains a reference to root, you always have a reference to the root of the tree. Every time you insert, you start with the root and then recursively traverse the tree until you find a suitable place to insert the new node. If you really must return the reference, then you most certainly should not be setting root to that new node!
Here is some pseudocode to help you out:
// Recursive function that inserts a value into a BST
function insert(node, value):
//Handles the case where you have no nodes in the tree, so root is null
if node is null:
node = new Node(value)
// If the value is lesser than the current node's value, we need to insert it
// somewhere in the right subtree
else if value < node.value:
if node.right is null:
// This node doesn't have a right child, so let's insert the new node here
node.right = new Node(value)
else:
// This node has a right child, so let's go further into this subtree to
// find the right place to insert the new node
insert(node.right, value)
// If the value is greater than the current node's value, we need to insert it
// somewhere in the left subtree
else if value > node.value:
if node.left is null:
// This node doesn't have a left child, so let's insert the new node here
node.left = new Node(value)
else:
// This node has a left child, so let's go further into this subtree to
// find the right place to insert the new node
insert(node.left, value)
else:
// A node with this value already exists so let's print out an erro
error("Node with that value already exists")
end function
Related
I decided to create a Binary Search Tree using java, and what I want to do is delete the Max element from the tree, so I created this part of code:
public Node<T> removeMax(Node<T> node) {
if (node == null)
return null;
if (node.right == null) {
Node<T> n = node;
node = null;
return n;
}
return removeMax(node.right);
}
The method returns the Max element, but it doesn't remove it from the tree. As you can see, I tried to remove it in this part:
Node<T> n = node;
node = null;
return n;
But, when I print the elements in the tree, it shows the "removed" ones too.
What am I doing wrong?
EDIT: What I'm really trying to do is delete the Max node and return it so I can now which one was deleted.
Now I noticed you wanted to know which node gets deleted. All you can do is to delete the maximum node and print the tree:
public void deleteMax(Node<T> root) {
Node<T> current = root;
while (current.right.right != null) {
current = current.right;
}
if(current.right.left!=null) {//max has 1 child to left
current.right=current.right.left;
}
else {//max has no child
current.right=null;
}
}
public String printInfix(Node<T> root) {//prints all the data in the tree in infix order
if(root==null) {
return "";
}
return printAll(root.left+" "+root.data+" "+printAll(root.right);
}
You want to delete a node in a binary search tree. So, basically what you want to do is to make it inaccessible. To do that, you have to nullify the reference to it, i.e, make its parent's corresponding pointer to it as null.
Change this:
if (node.right == null) {
Node<T> n = node;
node = null;
return n;
}
To this:
if (node.right.right == null) {
Node<T> n = node.right;
node.right = node.right.left;
return n;
}
Also you need to take care of the case when the root is the maximum element of the tree. So, if you have some reference to the root of BST, add this case before the above case:
if (node.right == null) {
Node<T> n = node;
referenceToTheRootOfBST = node.left;
return n;
}
If you don't have a reference to the root of the BST, what you can do is deep copy the left node of the root and then remove the left node. So, the above case changes to:
if (node.right == null) {
Node<T> n = node;
//I'll assume you don't call this function if root is the only element.
//if root is the only element.If that's the case, then just make the
//root null before calling this function.
Node<T> leftNode = node.left;
node.value = leftNode.value;
node.left = leftNode.left;
node.right = leftNode.right;
return n;
}
Another simple way to handle the case that root is the maximum element is to check it before actually calling this function. Simply check if root has a right node, and if it doesn't have it, reallocate the root reference.
Okay so I've been looking into this for days and everytime I think I've gotten it down I start writing code and I get to a point where I just can't figure out what exactly to do.
The tree isn't recursive, so I can follow everything really until I start trying to modify it so it uses lazy deletion instead of real deletion. (Right now it nulls out the node it deletes)
What I have managed to figure out:
I added a flag to the node class to set them as deleted
I've implemented a search method that works, it even seems to register if my nodes are deleted or not(lazily)
I know that the rest of the tree class should treat the nodes that are flagged as deleted such that they are not there.
What I don't know:
I've looked at MANY resources and some say all you need to do is set
the node's deleted flag to true. Does this mean that I don't have to
worry about the linking after their flag is set?
Is an appropriate way to do this very superficial? As in, just don't let the methods report that something is found if the flag is set to deleted even though the methods do find something?
In what method(s) should I change to use lazy deletion? Only the delete() method?
If I only change the delete method, how is this picked up by the other methods?
Does the search method look okay?
Here's the rest of the code so you can see what I'm using. I'm really frustrated because I honestly understand how to delete nodes completely way better then this stupid lazy deletion implementation. It's what they teach in the book! lol
Please help... :(
Search Method
So here's my search method:
public String search(E data){
Node<E> current = root;
String result = "";
while(current != null){
if(data.compareTo(current.e) < 0){
current = current.left;
}
else if (data.compareTo(current.e) > 0){
current = current.right;
}
else{
if (current.isDeleted == false){
return result += "Found node with matching data that is not deleted!";
}
else{
return result += "Found deleted data, not usable, continuing search\n";
}
}
}
return result += "Did not find non-deleted matching node!";
}
Tree Class
Tree Code (The real deletion method is commented out at the end so I could replace it with the lazy deletion):
package mybinarytreeexample;
public class MyBinaryTree> {
private Node<E> root = null;
public class Node<E> {
public boolean isDeleted = false;
public E e = null;
public Node<E> left = null;
public Node<E> right = null;
}
public boolean insert(E e) {
// if empty tree, insert a new node as the root node
// and assign the elementy to it
if (root == null) {
root = new Node();
root.e = e;
return true;
}
// otherwise, binary search until a null child pointer
// is found
Node<E> parent = null;
Node<E> child = root;
while (child != null) {
if (e.compareTo(child.e) < 0) {
parent = child;
child = child.left;
} else if (e.compareTo(child.e) > 0) {
parent = child;
child = child.right;
} else {
if(child.isDeleted){
child.isDeleted = false;
return true;
}
return false;
}
}
// if e < parent.e create a new node, link it to
// the binary tree and assign the element to it
if (e.compareTo(parent.e) < 0) {
parent.left = new Node();
parent.left.e = e;
} else {
parent.right = new Node();
parent.right.e = e;
}
return true;
}
public void inorder() {
System.out.print("inorder: ");
inorder(root);
System.out.println();
}
private void inorder(Node<E> current) {
if (current != null) {
inorder(current.left);
System.out.printf("%3s", current.e);
inorder(current.right);
}
}
public void preorder() {
System.out.print("preorder: ");
preorder(root);
System.out.println();
}
private void preorder(Node<E> current) {
if (current != null) {
System.out.printf("%3s", current.e);
preorder(current.left);
preorder(current.right);
}
}
public void postorder() {
System.out.print("postorder: ");
postorder(root);
System.out.println();
}
private void postorder(Node<E> current) {
if (current != null) {
postorder(current.left);
postorder(current.right);
System.out.printf("%3s", current.e);
}
}
public String search(E data){
Node<E> current = root;
String result = "";
while(current != null){
if(data.compareTo(current.e) < 0){
current = current.left;
}
else if (data.compareTo(current.e) > 0){
current = current.right;
}
else{
if (current.isDeleted == false){
return result += "Found node with matching data that is not deleted!";
}
else{
return result += "Found deleted data, not usable, continuing search\n";
}
}
}
return result += "Did not find non-deleted matching node!";
}
public boolean delete(E e) {
}
// an iterator allows elements to be modified, but can mess with
// the order if element not written with immutable key; it is better
// to use delete to remove and delete/insert to remove or replace a
// node
public java.util.Iterator<E> iterator() {
return new PreorderIterator();
}
private class PreorderIterator implements java.util.Iterator<E> {
private java.util.LinkedList<E> ll = new java.util.LinkedList();
private java.util.Iterator<E> pit= null;
// create a LinkedList object that uses a linked list of nodes that
// contain references to the elements of the nodes of the binary tree
// in preorder
public PreorderIterator() {
buildListInPreorder(root);
pit = ll.iterator();
}
private void buildListInPreorder(Node<E> current) {
if (current != null) {
ll.add(current.e);
buildListInPreorder(current.left);
buildListInPreorder(current.right);
}
}
// check to see if their is another node in the LinkedList
#Override
public boolean hasNext() {
return pit.hasNext();
}
// reference the next node in the LinkedList and return a
// reference to the element in the node of the binary tree
#Override
public E next() {
return pit.next();
}
#Override
public void remove() {
throw new UnsupportedOperationException("NO!");
}
}
}
// binary search until found or not in list
// boolean found = false;
// Node<E> parent = null;
// Node<E> child = root;
//
// while (child != null) {
// if (e.compareTo(child.e) < 0) {
// parent = child;
// child = child.left;
// } else if (e.compareTo(child.e) > 0) {
// parent = child;
// child = child.right;
// } else {
// found = true;
// break;
// }
// }
//
//
// if (found) {
// // if root only is the only node, set root to null
// if (child == root && root.left == null && root.right == null)
// root = null;
// // if leaf, remove
// else if (child.left == null && child.right == null) {
// if (parent.left == child)
// parent.left = null;
// else
// parent.right = null;
// } else
// // if the found node is not a leaf
// // and the found node only has a right child,
// // connect the parent of the found node (the one
// // to be deleted) to the right child of the
// // found node
// if (child.left == null) {
// if (parent.left == child)
// parent.left = child.right;
// else
// parent.right = child.right;
// } else {
// // if the found node has a left child,
// // the node in the left subtree with the largest element
// // (i. e. the right most node in the left subtree)
// // takes the place of the node to be deleted
// Node<E> parentLargest = child;
// Node<E> largest = child.left;
// while (largest.right != null) {
// parentLargest = largest;
// largest = largest.right;
// }
//
// // replace the lement in the found node with the element in
// // the right most node of the left subtree
// child.e = largest.e;
//
// // if the parent of the node of the largest element in the
// // left subtree is the found node, set the left pointer of the
// // found node to point to left child of its left child
// if (parentLargest == child)
// child.left = largest.left;
// else
// // otherwise, set the right child pointer of the parent of
// // largest element in the left subtreeto point to the left
// // subtree of the node of the largest element in the left
// // subtree
// parentLargest.right = largest.left;
// }
//
// } // end if found
//
// return found;
What changes is that your tree only grows in term of real space used, and never shrinks. This can be very useful if you choose a list as a data-structure to implement your tree, rather than the usual construct Node E {V value; E right; E; left}. I will come back on this later.
I've looked at MANY resources and some say all you need to do is set
the node's deleted flag to true. Does this mean that I don't have to
worry about the linking after their flag is set?
Yes, if by linking you mean node.left, node.right. Delete simply mark as deleted and that's it. It change nothing else, and it should not, because x.CompareTo(y) must be still working even if x or y are marked as deleted
Is an appropriate way to do this very superficial? As in, just don't
let the methods report that something is found if the flag is set to
deleted even though the methods do find something?
Well by definition of this method "something" means a node without the deleted flag. Anything with the deleted flag is "nothing" for the user of the tree.
what method(s) should I change to use lazy deletion? Only the delete()
method?
Of course not. You already changed the search method yourself. Let's take the isEmpty(). You should keep a counter of deleted nodes and one of total nodes. If they are equal the tree is empty. Otherwise the tree is not.
There is a small bug in your algorithm. When you insert and find out that you land on a deleted node, you just unmark that node. You must also set the value of the node. After all compareTo doesnt insure all fields are strictly equal, just that the objects are equivalent.
if(child.isDeleted){
child.isDeleted = false;
child.e = e; <---- missing
return true;
}
There might be others.
Side Note:
As said before one instance where this method is useful is a tree backed by an list (let's say array list). With this method the children of element at position i are at position 2*i+1 and 2*i+2. Usually when you delete a node p with children, you replace that node with the leftmost node q of the right subtree (or rightmost node in the left subtree). Here you can just mark p as deleted and swap the value of the deleted node and leftmost. Your array stays intact in memory
I am struggling to figure out how to code a recursive algorithm to count the number of leaves in a Binary Tree (not a complete tree). I get as far as traversing to the far most left leaf and don't know what to return from there. I am trying to get the count by loading the leaves into a list and getting the size of that list. This is probably a bad way to go about the count.
public int countLeaves ( ) {
List< Node<E> > leafList = new ArrayList< Node<E> >();
//BinaryTree<Node<E>> treeList = new BinaryTree(root);
if(root.left != null)
{
root = root.left;
countLeaves();
}
if(root.right != null)
{
root = root.right;
countLeaves();
}
if(root.left == null && root.right == null)
{
leafList.add(root);
}
return();
}
Elaborating on #dasblinkenlight idea. You want to recursively call a countleaves on root node & pass back the # to caller. Something on the following lines.
public int countLeaves() {
return countLeaves(root);
}
/**
* Recursively count all nodes
*/
private static int countLeaves (Node<E> node) {
if(node==null)
return 0;
if(node.left ==null && node.right == null)
return 1;
else {
return countLeaves(node.left) + countLeaves(node.right);
}
}
Edit: It appears, a similar problem was previously asked counting number of leaf nodes in binary tree
The problem with your implementation is that it does not restore the value of member variable root back to the state that it had prior to entering the method. You could do it by storing the value in a local variable, i.e.
Node<E> oldRoot = root;
... // your method goes here
root = oldRoot;
However, a better approach is to take Node<E> as an argument, rather than relying on a shared variable:
public int countLeaves() {
return countLeaves(root);
}
private static int countLeaves (Node<E> node) {
... // Do counting here
}
I have written a code to insert an element in a binary tree in java. Here are the functions to do the same:
public void insert(int data)
{
root = insert(root, data);
}
private Node insert(Node node, int data)
{
if (node == null)
node = new Node(data);
else
{
if (node.getRight() == null)
node.right = insert(node.right, data);
else
node.left = insert(node.left, data);
}
return node;
}
However when I traverse the tree, the answer I get is wrong. Here are the traversal functions (preorder):
public void preorder()
{
preorder(root);
}
private void preorder(Node r)
{
if (r != null)
{
System.out.print(r.getData() +" ");
preorder(r.getLeft());
preorder(r.getRight());
}
}
Okay so as suggested here's the definition for the Node class:
public class Node {
public int data;
public Node left, right;
/* Constructor */
public Node() {
left = null;
right = null;
data = 0;
}
/* Constructor */
public Node(int d, Node l, Node r) {
data = d;
left = l;
right = r;
}
//Constructor
public Node(int d) {
data = d;
}
/* Function to set link to next Node */
public void setLeft(Node l) {
left = l;
}
/* Function to set link to previous Node */
public void setRight(Node r) {
right = r;
}
/* Function to set data to current Node */
public void setData(int d) {
data = d;
}
/* Function to get link to next node */
public Node getLeft() {
return left;
}
/* Function to get link to previous node */
public Node getRight() {
return right;
}
/* Function to get data from current Node */
public int getData() {
return data;
}
}
I have re-checked the algorithm for traversal many times, and it's working perfectly. I believe the problem is in the insertion algorithm. Any suggestions?
If I understood correctly, you want to fill your binary tree in "layers". E.g. you want to put something into depth 4 only if depth 3 is "full binary tree".
Then the problem is whole logic of your insert algorithm that is DFS-based. In other words it inserts elements deeper and deeper on the one side instead of building full binary tree on both sides.
If you look closer to your insert algorithm you will see that once you skip "right" subtree, you will never return to it - even if the "left" subtree is already full binary tree. That leads to the tree that will be growing deeper and deeper on the left side but not growing on the right side.
Speaking in programming language. You do:
(node.right != null) && (node.left != null) => insert (node.left)
but you can't do this (start inserting node.left). What if node.left has both children and node.right has no children? You will attempt to insert to the left even you should do it in node.right.
So what you really need to do insertion BFS-based. That means you will traverse the tree for insertion "in layers". Queue should be your new friend here:-) (not the stack/recursion):
public void insert(int data) {
if (root == null) {
root = new Node(data);
return;
}
Queue<Node> nodesToProcess = new LinkedList<>();
nodesToProcess.add(root);
while (true) {
Node actualNode = nodesToProcess.poll();
// Left child has precedence over right one
if (actualNode.left == null) {
actualNode.left = new Node(data);
return;
}
if (actualNode.right == null) {
actualNode.right = new Node(data);
return;
}
// I have both children set, I will process them later if needed
nodesToProcess.add(actualNode.left);
nodesToProcess.add(actualNode.right);
}
}
Your method returns given node, but your method has to return inserted node which is node.right or node.left
I'm a little confused why my code doesn't insert nodes past the first one. say I wanted to insert (5,4) after (7,2) using the code below; in this case the very first condition is triggered, isVertical and p.x() < node.point.x(). The way I see it since node.left is null, I exit the loop and since node holds a reference node.left based on the latest assignment, I should be able to use node to insert a new tree leaf. Am I not seeing this right? Is node not really a reference to node.left? Sorry if this is a dumb question, I am still just a little shaky with references.
public void insert(Point2D p) {
if (p == null) {
throw new java.lang.NullPointerException();
}
if (size == 0) {
root = new Node(p);
size++;
return;
}
Node node = root;
while (node != null) {
// sink
if (node.isVertical()) {
if (p.x() < node.point.x()) {
node = node.left; // go left
} else {
node = node.right; // go right
}
} else if (node.isHorizontal()) {
if (p.y() < node.point.y()) {
node = node.left; // go left
} else {
node = node.right; // go right
}
}
}
node = new Node(p);
}
You're just assigning new Node(p) to a local variable node, whose value is lost as soon as the function returns. To change the existing tree, your assignment should have the form node.left = new Node(p); or node.right = new Node(p).