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Why am I getting an incorrect answer while trying to find the closest value in a BST given a target value?

There is only one difference between the correct answer and my answer, and that is, I am traversing the entire tree instead of comparing the target with the node value and eliminating one-half of the tree in each recursion. Please help me with the explanation. Thanks.
My code:
import java.util.*;
class Program {
public static int findClosestValueInBst(BST tree, int target) {
//int closest = Integer.MAX_VALUE;
// int val = 0;
int vl = findClosestValueInBst1(tree, target, tree.value);
return vl;
}
public static int findClosestValueInBst1(BST tree, int target, int val) {
// System.out.println((closest + " " + Math.abs(tree.value - target)));
//c = closest;
if(( Math.abs(target - tree.value)) < ( Math.abs(target - val))){
System.out.println(val);
val = tree.value;
}
if(tree.left != null){
return findClosestValueInBst1(tree.left, target, val);
}
if(tree.right != null){
return findClosestValueInBst1(tree.right, target, val);
}
return val;
}
static class BST {
public int value;
public BST left;
public BST right;
public BST(int value) {
this.value = value;
}
}
}
Question tree- Root =10,
Nodes-> [10,15,22,13,14,5,5,2,1],
Target: 12,
My output: 10,
Correct answer: 13,
import java.util.*;
class Program {
public static int findClosestValueInBst(BST tree, int target) {
//int closest = Integer.MAX_VALUE;
// int val = 0;
int vl = findClosestValueInBst1(tree, target, tree.value);
return vl;
}
public static int findClosestValueInBst1(BST tree, int target, int val) {
// System.out.println((closest + " " + Math.abs(tree.value - target)));
//c = closest;
if(( Math.abs(target - tree.value)) < ( Math.abs(target - val))){
System.out.println(val);
val = tree.value;
}
if( target < tree.value && tree.left != null){
return findClosestValueInBst1(tree.left, target, val);
} else
if(target > tree.value && tree.right != null){
return findClosestValueInBst1(tree.right, target, val);
} else
return val;
}
static class BST {
public int value;
public BST left;
public BST right;
public BST(int value) {
this.value = value;
}
}
}

The tree looks like this:
10
/\
5 15
/ /\
2 13 22
/ \
1 14
Your code is not actually traversing the whole tree. This code:
if(tree.left != null){
return findClosestValueInBst1(tree.left, target, val);
}
if(tree.right != null){
return findClosestValueInBst1(tree.right, target, val);
}
return val;
checks the left subtree if it exists (and ignores the right subtree). Otherwise, check the right subtree if it exists. Otherwise stop the recursion. This is because once you reach a return statement, the entire method stops there, and the lines after that do not get executed.
So your code always prefers the left subtree without taking into account what number the node actually stores. So right off the bat, you went to the wrong direction - you are looking for 13, and the current node is 10, a closer value is gotta be bigger than 10, i.e. in the right subtree.
An implementation that actually traverses the whole tree would be something like:
public static int findClosestValueInBst(BST tree, int target) { // no need for the val argument!
int leftClosest = tree.value;
int rightClosest = tree.value;
if(tree.left != null){
leftClosest = findClosestValueInBst1(tree.left, target);
}
if(tree.right != null){
rightClosest = findClosestValueInBst1(tree.right, target);
}
if (target - leftClosest < rightClosest - target) {
return leftClosest;
} else {
return rightClosest;
}
}
But why bother when you can do it more quickly? :)

Printing values in a tree structure in Java?

I am trying to write this code to print the given user inputs in a tree structure that follows
x
x x
x x
but it does not output that way.
I am getting the output as
x
x
x
This is the function I have written that gets and prints:
private void inOrder(Node n)
{
if(n == null) // recursion ends when node is null
return;
{
inOrder(n.left);
System.out.println(n.data);
inOrder(n.right);
}
}
public void printInorder()
{
inOrder(root);
}

This approach runs into trouble because any calls to println() preclude printing further nodes on a line. Using a level order traversal/BFS will enable you to call println() to move to the next line only when all nodes on a given tree level have already been printed.
The bigger difficulty lies in keeping track of the horizontal placement of each node in a level. Doing this properly involves considering the depth, length of the node data and any empty children. If you can, consider printing your tree with depth increasing from left to right, similar to the unix command tree, rather than top-down, which simplifies the algorithm.
Here's a proof-of-concept for a top-down print. Spacing formulas are from this excellent post on this very topic. The strategy I used is to run a BFS using a queue, storing nodes (and null placeholders) in a list per level. Once the end of a level is reached, spacing is determined based on the number of nodes on a level, which is 2n-1, and printed. A simplifying assumption is that node data width is 1.
import java.util.*;
import static java.lang.System.out;
public class Main {
static void printLevelOrder(Node root) {
LinkedList<QItem> queue = new LinkedList<>();
ArrayList<Node> level = new ArrayList<>();
int depth = height(root);
queue.add(new QItem(root, depth));
for (;;) {
QItem curr = queue.poll();
if (curr.depth < depth) {
depth = curr.depth;
for (int i = (int)Math.pow(2, depth) - 1; i > 0; i--) {
out.print(" ");
}
for (Node n : level) {
out.print(n == null ? " " : n.val);
for (int i = (int)Math.pow(2, depth + 1); i > 1; i--) {
out.print(" ");
}
}
out.println();
level.clear();
if (curr.depth <= 0) {
break;
}
}
level.add(curr.node);
if (curr.node == null) {
queue.add(new QItem(null, depth - 1));
queue.add(new QItem(null, depth - 1));
}
else {
queue.add(new QItem(curr.node.left, depth - 1));
queue.add(new QItem(curr.node.right, depth - 1));
}
}
}
static int height(Node root) {
return root == null ? 0 : 1 + Math.max(
height(root.left), height(root.right)
);
}
public static void main(String[] args) {
printLevelOrder(
new Node<Integer>(
1,
new Node<Integer>(
2,
new Node<Integer>(
4,
new Node<Integer>(7, null, null),
new Node<Integer>(8, null, null)
),
null
),
new Node<Integer>(
3,
new Node<Integer>(
5,
new Node<Integer>(9, null, null),
null
),
new Node<Integer>(
6,
null,
new Node<Character>('a', null, null)
)
)
)
);
}
}
class Node<T> {
Node left;
Node right;
T val;
public Node(T val, Node left, Node right) {
this.left = left;
this.right = right;
this.val = val;
}
}
class QItem {
Node node;
int depth;
public QItem(Node node, int depth) {
this.node = node;
this.depth = depth;
}
}
Output:
1
2 3
4 5 6
7 8 9 a
Try it!

You have a problem to print the way you want.
An In order will print left, current and right. These are in different levels in the tree. As soon you print a down level you can't print the current one above because it was already printed.
Also don't forget the println will print that string and give a new line after.
To have a fancy design, you probably need to do some fancy engineering to align them perfectly, something like this:
You need a Queue for the nodes to visit.
printNode(Node root, queueWithNodesAndStartAndEnd, start, end)
print me at the middle of start and end
put my left child in the queue with start = myStart and end = middle of my start and my end
put my right child in the queue with start = middle of my start and my end and end = my end
pop (get and remove) first element from queue
if not empty print popped node with start and end provided
I know this is pseudo-code but you should be able to implement it.

This is the nicest solution I've seen: https://stackoverflow.com/a/42449385/9319615
Here is my code snippet leveraging it. This class will run as is.
class Node {
final int value;
Node left;
Node right;
Node(int value) {
this.value = value;
right = null;
left = null;
}
public void print() {
print("", this, false);
}
private void print(String prefix, Node n, boolean isLeft) {
if (n != null) {
System.out.println(prefix + (isLeft ? "|-- " : "\\-- ") + n.value);
print(prefix + (isLeft ? "| " : " "), n.left, true);
print(prefix + (isLeft ? "| " : " "), n.right, false);
}
}
}
class BinaryTree {
Node root;
private Node addRecursive(Node current, int value) {
if (current == null) {
return new Node(value);
}
if (value < current.value) {
current.left = addRecursive(current.left, value);
} else if (value > current.value) {
current.right = addRecursive(current.right, value);
} else {
// value already exists
return current;
}
return current;
}
public void add(int value) {
root = addRecursive(root, value);
}
public void traverseInOrder(Node node) {
if (node != null) {
traverseInOrder(node.left);
System.out.print(" " + node.value);
traverseInOrder(node.right);
}
}
}
public class Main {
public static void main(String[] args) {
BinaryTree bt = new BinaryTree();
bt.add(6);
bt.add(4);
bt.add(8);
bt.add(3);
bt.add(5);
bt.add(7);
bt.add(9);
System.out.println("Print in order->");
bt.traverseInOrder(bt.root);
System.out.println("\n\nPrint Tree Structure");
bt.root.print();
}
}
Output example

convert binary to decimal using single linked list and recursive method in java

How we can convert binary to decimal using single linked list and recursive method in java? :-s
Ex:
Input: 1->0->0->NULL
Output: 4

I can think of two ways to solve it:
1- If length of list is known:
// To find length of list
public int length(Node head) {
int count = 0;
while(head != null) {
count++;
head = head.next;
}
return count;
}
public static int convertWhenLengthIsKnown(Node head, int len) {
int sum = 0;
if(head.next != null) {
sum = convertWhenLengthIsKnown(head.next, len-1);
}
return sum + head.data * (int)Math.pow(2,len);
}
// Call this function as below:
convertWhenLengthIsKnown(head, length(head)-1);
If we don't want to calculate length, then we can have a sum variable which is globally accessible,
private static int sum = 0;
public static int convert(Node head,int i) {
if(head.next != null) {
i = convert(head.next, i);
}
sum += head.data * (int)Math.pow(2,i);
return i+1;
}
// Call this function as below:
convert(head,0);
Below is the Node class:
class Node {
int data;
Node next;
Node(int data) {
this.data = data;
}
}
Hope It helps you.

It's a bit tricky recursively because you need to scale "earlier" values depending on the list length, simplest way seems to be to hand down the result "so far" via an additional parameter.
class Node {
int value;
Node next;
int toDecimal(int resultSoFar) {
int resultSoFar = 2 * resultSoFar + value;
return next == null ? resultSoFar : toDecimal(resultSoFar);
}
int toDecimal() {
toDecimal(0);
}
}

Try this (Without the null at the end of the list) :
public Integer binToDec(LinkedList<Integer> list){
Double size = (double) list.size();
if(size == 0D) return 0;
Integer number = list.getFirst();
list.removeFirst();
if(number != 0) return binToDec(list) + number*new Double(Math.pow(2D, size - 1)).intValue();
else return binToDec(list);
}
Be aware that it will clear the list.

As you did not tell anything about your linked list, I assume it's similar to the Node class from pbajpai21.
Without knowing the length of the chain you could with each level shift the value one position to the left.
the class for each node in the list
class Node {
int digit;
Node child;
Node(int data) {
this.digit = data;
}
public int getDigit() {
return digit;
}
public void setChild(Node child) {
this.child = child;
}
public Node getChild() {
return child;
}
}
the class for demonstration
public class Bits {
public static void main(String[] args) {
int[] ints = new int[] {1, 0, 1, 0, 1, 0};
Node parent = new Node(ints[0]);
Node root = parent;
for (int i = 1; i < ints.length; i++) {
Node child = new Node(ints[i]);
parent.setChild(child);
parent = child;
}
long value = computeValue(0, root);
System.out.println();
System.out.println("value = " + value);
}
private static long computeValue(long parentValue, Node node) {
if (node == null) {
return parentValue;
}
// only to print the current digit
System.out.print(node.getDigit());
long value = (parentValue << 1) + node.getDigit();
return computeValue(value, node.getChild());
}
}
output
101010
value = 42

Recursively find nth to last element in linked list

I'm practicing basic data structure stuff and I'm having some difficulties with recursion. I understand how to do this through iteration but all of my attempts to return the nth node from the last of a linked list via recursion result in null. This is my code so far:
public static int i = 0;
public static Link.Node findnthToLastRecursion(Link.Node node, int pos) {
if(node == null) return null;
else{
findnthToLastRecursion(node.next(), pos);
if(++i == pos) return node;
return null;
}
Can anyone help me understand where I'm going wrong here?
This is my iterative solution which works fine, but I'd really like to know how to translate this into recursion:
public static Link.Node findnthToLast(Link.Node head, int n) {
if (n < 1 || head == null) {
return null;
}
Link.Node pntr1 = head, pntr2 = head;
for (int i = 0; i < n - 1; i++) {
if (pntr2 == null) {
return null;
} else {
pntr2 = pntr2.next();
}
}
while (pntr2.next() != null) {
pntr1 = pntr1.next();
pntr2 = pntr2.next();
}
return pntr1;
}

You need to go to the end and then count your way back, make sure to pass back the node each time its passed back. I like one return point
public static int i = 0;
public static Link.Node findnthToLastRecursion(Link.Node node, int pos) {
Link.Node result = node;
if(node != null) {
result = findnthToLastRecursion(node.next, pos);
if(i++ == pos){
result = node;
}
}
return result;
}
Working example outputs 7 as 2 away from the 9th and last node:
public class NodeTest {
private static class Node<E> {
E item;
Node<E> next;
Node<E> prev;
Node(Node<E> prev, E element, Node<E> next) {
this.item = element;
this.next = next;
this.prev = prev;
}
}
/**
* #param args
*/
public static void main(String[] args) {
Node first = null;
Node prev = null;
for (int i = 0; i < 10; i++) {
Node current = new Node(prev, Integer.toString(i),null);
if(i==0){
first = current;
}
if(prev != null){
prev.next = current;
}
prev = current;
}
System.out.println( findnthToLastRecursion(first,2).item);
}
public static int i = 0;
public static Node findnthToLastRecursion(Node node, int pos) {
Node result = node;
if (node != null) {
result = findnthToLastRecursion(node.next, pos);
if (i++ == pos) {
result = node;
}
}
return result;
}
}

No need for static variables.
public class List {
private Node head = null;
// [...] Other methods
public Node findNthLastRecursive(int nth) {
if (nth <= 0) return null;
return this.findNthLastRecursive(this.head, nth, new int[] {0});
}
private Node findNthLastRecursive(Node p, int nth, int[] pos) {
if (p == null) {
return null;
}
Node n = findNthLastRecursive(p.next, nth, pos);
pos[0]++;
if (pos[0] == nth) {
n = p;
}
return n;
}
}

You can do this a couple of ways:
recurse through the list once to find the list length, then write a recursive method to return the kth element (a much easier problem).
use an auxiliary structure to hold the result plus the remaining length; this essentially replaces the two recursions of the first option with a single recursion:
static class State {
Link.Node result;
int trailingLength;
}
public static Link.Node findnthToLastRecursion(Link.Node node, int pos) {
if(node == null) return null;
State state = new State();
findnthToLastRecursion(node, pos, state);
return state.result;
}
private static void findnthToLastRecursion(Link.Node node, int pos, State state) {
if (node == null) {
state.trailingLength = 0;
} else {
findnthToLastRecursion(node.next(), state);
if (pos == state.trailingLength) {
state.result = node;
}
++state.trailingLength;
}
}

I misunderstood the question. Here is an answer based on your iterative solution:
public static Link.Node findnthToLast(Link.Node head, int n) {
return findnthToLastHelper(head, head, n);
}
private static Link.Node findnthToLastHelper(Link.Node head, Link.Node end, int n) {
if ( end == null ) {
return ( n > 0 ? null : head);
} elseif ( n > 0 ) {
return findnthToLastHelper(head, end.next(), n-1);
} else {
return findnthToLastHelper(head.next(), end.next(), 0);
}
}

actually you don't need to have public static int i = 0; . for utill method the pos is :
pos = linked list length - pos from last + 1
public static Node findnthToLastRecursion(Node node, int pos) {
if(node ==null){ //if null then return null
return null;
}
int length = length(node);//find the length of the liked list
if(length < pos){
return null;
}
else{
return utill(node, length - pos + 1);
}
}
private static int length(Node n){//method which finds the length of the linked list
if(n==null){
return 0;
}
int count = 0;
while(n!=null){
count++;
n=n.next;
}
return count;
}
private static Node utill(Node node, int pos) {
if(node == null) {
return null;
}
if(pos ==1){
return node;
}
else{
return utill(node.next, pos-1);
}
}
Here node.next is the next node. I am directly accessing the next node rather than calling the next() method. Hope it helps.

This cheats (slightly) but it looks good.
public class Test {
List<String> list = new ArrayList<> (Arrays.asList("Zero","One","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten"));
public static String findNthToLastUsingRecursionCheatingALittle(List<String> list, int n) {
int s = list.size();
return s > n
// Go deeper!
? findNthToLastUsingRecursionCheatingALittle(list.subList(1, list.size()), n)
// Found it.
: s == n ? list.get(0)
// Too far.
: null;
}
public void test() {
System.out.println(findNthToLastUsingRecursionCheating(list,3));
}
public static void main(String args[]) {
new Test().test();
}
}
It prints:
Eight
which I suppose is correct.
I have use List instead of some LinkedList variant because I do not want to reinvent anything.

int nthNode(struct Node* head, int n)
{
if (head == NULL)
return 0;
else {
int i;
i = nthNode(head->left, n) + 1;
printf("=%d,%d,%d\n", head->data,i,n);
if (i == n)
printf("%d\n", head->data);
}
}

public class NthElementFromLast {
public static void main(String[] args) {
List<String> list = new LinkedList<>();
Stream.of("A","B","C","D","E").forEach(s -> list.add(s));
System.out.println(list);
System.out.println(getNthElementFromLast(list,2));
}
private static String getNthElementFromLast(List list, int positionFromLast) {
String current = (String) list.get(0);
int index = positionFromLast;
ListIterator<String> listIterator = list.listIterator();
while(positionFromLast>0 && listIterator.hasNext()){
positionFromLast--;
current = listIterator.next();
}
if(positionFromLast != 0) {
return null;
}
String nthFromLast = null;
ListIterator<String> stringListIterator = list.listIterator();
while(listIterator.hasNext()) {
current = listIterator.next();
nthFromLast = stringListIterator.next();
}
return nthFromLast;
}
}
This will find Nth element from last.

My approach is simple and straight,you can change the array size depending upon your requirement:
int pos_from_tail(node *k,int n)
{ static int count=0,a[100];
if(!k) return -1;
else
pos_from_tail(k->next,n);
a[count++]=k->data;
return a[n];
}

You'll have make slight changes in the code:
public static int i = 0;
public static Link.Node findnthToLastRecursion(Link.Node node, int pos) {
if(node == null) return null;
else{
**Link.Node temp = findnthToLastRecursion(node.next(), pos);
if(temp!=null)
return temp;**
if(++i == pos) return node;
return null;
}
}

Binary Heap Implemented via a Binary Tree Structure

For an assignment, we were instructed to create a priority queue implemented via a binary heap, without using any built-in classes, and I have done so successfully by using an array to store the queued objects. However, I'm interested in learning how to implement another queue by using an actual tree structure, but in doing so I've run across a bit of a problem.
How would I keep track of the nodes on which I would perform insertion and deletion? I have tried using a linked list, which appends each node as it is inserted - new children are added starting from the first list node, and deleted from the opposite end. However, this falls apart when elements are rearranged in the tree, as children are added at the wrong position.
Edit: Perhaps I should clarify - I'm not sure how I would be able to find the last occupied and first unoccupied leaves. For example, I would always be able to tell the last inserted leaf, but if I were to delete it, how would I know which leaf to delete when I next remove the item? The same goes for inserting - how would I know which leaf to jump to next after the current leaf has both children accounted for?

A tree implementation of a binary heap uses a complete tree [or almost full tree: every level is full, except the deepest one].
You always 'know' which is the last occupied leaf - where you delete from [and modifying it is O(logn) after it changed so it is not a problem], and you always 'know' which is the first non-occupied leaf, in which you add elements to [and again, modifying it is also O(logn) after it changed].
The algorithm idea is simple:
insert: insert element to the first non-occupied leaf, and use heapify [sift up] to get this element to its correct place in the heap.
delete_min: replace the first element with the last occupied leaf, and remove the last occupied leaf. then, heapify [sift down] the heap.
EDIT: note that delete() can be done to any element, and not only the head, however - finding the element you want to replace with the last leaf will be O(n), which will make this op expensive. for this reason, the delete() method [besides the head], is usually not a part of the heap data structure.

I really wanted to do this for almost a decade.Finally sat down today and wrote it.Anyone who wants it can use it.I got inspired by Quora founder to relearn Heap.Apparently he was asked how would you find K near points in a set of n points in his Google phone screen.Apparently his answer was to use a Max Heap and to store K values and remove the maximum element after the size of the heap exceeds K.The approach is pretty simple and the worst case is nlog K which is better than n^2 in most sorting cases.Here is the code.
import java.util.ArrayList;
import java.util.List;
/**
* #author Harish R
*/
public class HeapPractise<T extends Comparable<T>> {
private List<T> heapList;
public List<T> getHeapList() {
return heapList;
}
public void setHeapList(List<T> heapList) {
this.heapList = heapList;
}
private int heapSize;
public HeapPractise() {
this.heapList = new ArrayList<>();
this.heapSize = heapList.size();
}
public void insert(T item) {
if (heapList.size() == 0) {
heapList.add(item);
} else {
siftUp(item);
}
}
public void siftUp(T item) {
heapList.add(item);
heapSize = heapList.size();
int currentIndex = heapSize - 1;
while (currentIndex > 0) {
int parentIndex = (int) Math.floor((currentIndex - 1) / 2);
T parentItem = heapList.get(parentIndex);
if (parentItem != null) {
if (item.compareTo(parentItem) > 0) {
heapList.set(parentIndex, item);
heapList.set(currentIndex, parentItem);
currentIndex = parentIndex;
continue;
}
}
break;
}
}
public T delete() {
if (heapList.size() == 0) {
return null;
}
if (heapList.size() == 1) {
T item = heapList.get(0);
heapList.remove(0);
return item;
}
return siftDown();
}
public T siftDown() {
T item = heapList.get(0);
T lastItem = heapList.get(heapList.size() - 1);
heapList.remove(heapList.size() - 1);
heapList.set(0, lastItem);
heapSize = heapList.size();
int currentIndex = 0;
while (currentIndex < heapSize) {
int leftIndex = (2 * currentIndex) + 1;
int rightIndex = (2 * currentIndex) + 2;
T leftItem = null;
T rightItem = null;
int currentLargestItemIndex = -1;
if (leftIndex <= heapSize - 1) {
leftItem = heapList.get(leftIndex);
}
if (rightIndex <= heapSize - 1) {
rightItem = heapList.get(rightIndex);
}
T currentLargestItem = null;
if (leftItem != null && rightItem != null) {
if (leftItem.compareTo(rightItem) >= 0) {
currentLargestItem = leftItem;
currentLargestItemIndex = leftIndex;
} else {
currentLargestItem = rightItem;
currentLargestItemIndex = rightIndex;
}
} else if (leftItem != null && rightItem == null) {
currentLargestItem = leftItem;
currentLargestItemIndex = leftIndex;
}
if (currentLargestItem != null) {
if (lastItem.compareTo(currentLargestItem) >= 0) {
break;
} else {
heapList.set(currentLargestItemIndex, lastItem);
heapList.set(currentIndex, currentLargestItem);
currentIndex = currentLargestItemIndex;
continue;
}
}
}
return item;
}
public static void main(String[] args) {
HeapPractise<Integer> heap = new HeapPractise<>();
for (int i = 0; i < 32; i++) {
heap.insert(i);
}
System.out.println(heap.getHeapList());
List<Node<Integer>> nodeArray = new ArrayList<>(heap.getHeapList()
.size());
for (int i = 0; i < heap.getHeapList().size(); i++) {
Integer heapElement = heap.getHeapList().get(i);
Node<Integer> node = new Node<Integer>(heapElement);
nodeArray.add(node);
}
for (int i = 0; i < nodeArray.size(); i++) {
int leftNodeIndex = (2 * i) + 1;
int rightNodeIndex = (2 * i) + 2;
Node<Integer> node = nodeArray.get(i);
if (leftNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> leftNode = nodeArray.get(leftNodeIndex);
node.left = leftNode;
}
if (rightNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> rightNode = nodeArray.get(rightNodeIndex);
node.right = rightNode;
}
}
BTreePrinter.printNode(nodeArray.get(0));
}
}
public class Node<T extends Comparable<?>> {
Node<T> left, right;
T data;
public Node(T data) {
this.data = data;
}
}
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class BTreePrinter {
public static <T extends Comparable<?>> void printNode(Node<T> root) {
int maxLevel = BTreePrinter.maxLevel(root);
printNodeInternal(Collections.singletonList(root), 1, maxLevel);
}
private static <T extends Comparable<?>> void printNodeInternal(
List<Node<T>> nodes, int level, int maxLevel) {
if (nodes.isEmpty() || BTreePrinter.isAllElementsNull(nodes))
return;
int floor = maxLevel - level;
int endgeLines = (int) Math.pow(2, (Math.max(floor - 1, 0)));
int firstSpaces = (int) Math.pow(2, (floor)) - 1;
int betweenSpaces = (int) Math.pow(2, (floor + 1)) - 1;
BTreePrinter.printWhitespaces(firstSpaces);
List<Node<T>> newNodes = new ArrayList<Node<T>>();
for (Node<T> node : nodes) {
if (node != null) {
String nodeData = String.valueOf(node.data);
if (nodeData != null) {
if (nodeData.length() == 1) {
nodeData = "0" + nodeData;
}
}
System.out.print(nodeData);
newNodes.add(node.left);
newNodes.add(node.right);
} else {
newNodes.add(null);
newNodes.add(null);
System.out.print(" ");
}
BTreePrinter.printWhitespaces(betweenSpaces);
}
System.out.println("");
for (int i = 1; i <= endgeLines; i++) {
for (int j = 0; j < nodes.size(); j++) {
BTreePrinter.printWhitespaces(firstSpaces - i);
if (nodes.get(j) == null) {
BTreePrinter.printWhitespaces(endgeLines + endgeLines + i
+ 1);
continue;
}
if (nodes.get(j).left != null)
System.out.print("//");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(i + i - 1);
if (nodes.get(j).right != null)
System.out.print("\\\\");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(endgeLines + endgeLines - i);
}
System.out.println("");
}
printNodeInternal(newNodes, level + 1, maxLevel);
}
private static void printWhitespaces(int count) {
for (int i = 0; i < 2 * count; i++)
System.out.print(" ");
}
private static <T extends Comparable<?>> int maxLevel(Node<T> node) {
if (node == null)
return 0;
return Math.max(BTreePrinter.maxLevel(node.left),
BTreePrinter.maxLevel(node.right)) + 1;
}
private static <T> boolean isAllElementsNull(List<T> list) {
for (Object object : list) {
if (object != null)
return false;
}
return true;
}
}
Please note that BTreePrinter is a code I took somewhere in Stackoverflow long back and I modified to use with 2 digit numbers.It will be broken if we move to 3 digit numbers and it is only for simple understanding of how the Heap structure looks.A fix for 3 digit numbers is to keep everything as multiple of 3.
Also due credits to Sesh Venugopal for wonderful tutorial on Youtube on Heap data structure

public class PriorityQ<K extends Comparable<K>> {
private class TreeNode<T extends Comparable<T>> {
T val;
TreeNode<T> left, right, parent;
public String toString() {
return this.val.toString();
}
TreeNode(T v) {
this.val = v;
left = null;
right = null;
}
public TreeNode<T> insert(T val, int position) {
TreeNode<T> parent = findNode(position/2);
TreeNode<T> node = new TreeNode<T>(val);
if(position % 2 == 0) {
parent.left = node;
} else {
parent.right = node;
}
node.parent = parent;
heapify(node);
return node;
}
private void heapify(TreeNode<T> node) {
while(node.parent != null && (node.parent.val.compareTo(node.val) < 0)) {
T temp = node.val;
node.val = node.parent.val;
node.parent.val = temp;
node = node.parent;
}
}
private TreeNode<T> findNode(int pos) {
TreeNode<T> node = this;
int reversed = 1;
while(pos > 0) {
reversed <<= 1;
reversed |= (pos&1);
pos >>= 1;
}
reversed >>= 1;
while(reversed > 1) {
if((reversed & 1) == 0) {
node = node.left;
} else {
node = node.right;
}
reversed >>= 1;
}
return node;
}
public TreeNode<T> remove(int pos) {
if(pos <= 1) {
return null;
}
TreeNode<T> last = findNode(pos);
if(last.parent.right == last) {
last.parent.right = null;
} else {
last.parent.left = null;
}
this.val = last.val;
bubbleDown();
return null;
}
public void bubbleDown() {
TreeNode<T> node = this;
do {
TreeNode<T> left = node.left;
TreeNode<T> right = node.right;
if(left != null && right != null) {
T max = left.val.compareTo(right.val) > 0 ? left.val : right.val;
if(max.compareTo(node.val) > 0) {
if(left.val.equals(max)) {
left.val = node.val;
node.val = max;
node = left;
} else {
right.val = node.val;
node.val = max;
node = right;
}
} else {
break;
}
} else if(left != null) {
T max = left.val;
if(left.val.compareTo(node.val) > 0) {
left.val = node.val;
node.val = max;
node = left;
} else {
break;
}
} else {
break;
}
} while(true);
}
}
private TreeNode<K> root;
private int position;
PriorityQ(){
this.position = 1;
}
public void insert(K val) {
if(val == null) {
return;
}
if(root == null) {
this.position = 1;
root = new TreeNode<K>(val);
this.position++;
return ;
}
root.insert(val, position);
position++;
}
public K remove() {
if(root == null) {
return null;
}
K val = root.val;
root.remove(this.position-1);
this.position--;
if(position == 1) {
root = null;
}
return val;
}
public static void main(String[] args) {
PriorityQ<Integer> q = new PriorityQ<>();
System.out.println(q.remove());
q.insert(1);
q.insert(11);
q.insert(111);
q.insert(1111);
q.remove();
q.remove();
q.remove();
q.remove();
q.insert(2);
q.insert(4);
}
}