Modify Binary Tree by replacing all nodes at even and odd levels by their nearest even or odd perfect squares respectively

Given a Binary Tree consisting of N nodes, the task is to replace all the nodes that are present at even-numbered levels in a Binary Tree with their nearest even perfect square and replace nodes at odd-numbered levels with their nearest odd perfect square.

Examples:

Input:           5
                  /    \
                3       2
                      /    \ 
                    16    19

Output:        9
                  /    \
                4       4
                      /    \ 
                    9     25

Explanation: 
Level 1: Nearest odd perfect square to 5 is 9.
Level 2: Nearest even perfect square to 3 and 2 are 4.
Level 3: Nearest odd perfect square to 16 is 9 and to 19 is 25.

Input:          45
                  /    \
               65     32
                      /     
                    89    

Output:       49
                  /    \
                64    36
                      /    
                   81

Explanation: 
Level 1: Nearest odd perfect square to 45 is 49.
Level 2: Nearest even perfect square to 65 and 32 are 64 and 36 respectively.
Level 3: Nearest odd perfect square to 89 is 81.

Approach: The given problem can be solved using Level Order Traversal. Follow the steps below to solve the problem:

1. Initialize a queue, say q. 
2. Push root into the queue.
3. Loop while the queue is not empty
   • Store the current node in a variable, say temp_node.
   • If the current node value is a perfect square, check the following:
     • If the value of level is odd, and the value of the current node is even, find the nearest odd perfect square and update temp_node?data.
     • If the value of level is even and the value of the current node is odd, find the nearest even perfect square and update temp_node?data.
   • Otherwise, find the nearest odd perfect square if the value of level is odd or the closest even perfect square if the value of level is even. Update temp_node?data.
   • Print temp_node?data.
   • Enqueue temp_node’s children (first left, then right child) to q, if exists.
   • Dequeue current node from q.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Structure of a
// Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};

// Function to replace all nodes
// at even and odd levels with their
// nearest even or odd perfect squares
void LevelOrderTraversal(Node* root)
{
    // Base Case
    if (root == NULL)
        return;

    // Create an empty queue
    // for level order traversal
    queue<Node*> q;

    // Enqueue root
    q.push(root);

    // Initialize height
    int lvl = 1;

    // Iterate until queue is not empty
    while (q.empty() == false) {

        // Store the size
        // of the queue
        int n = q.size();

        // Traverse in range [1, n]
        for (int i = 0; i < n; i++) {

            // Store the current node
            Node* node = q.front();

            // Store its square root
            double num = sqrt(node->data);
            int x1 = floor(num);
            int x2 = ceil(num);

            // Check if it is a perfect square
            if (x1 == x2) {

                // If level is odd and value is even,
                // find the closest odd perfect square
                if ((lvl & 1) && !(x1 & 1)) {

                    int num1 = x1 - 1, num2 = x1 + 1;

                    node->data
                        = (abs(node->data - num1 * num1)
                           < abs(node->data - num2 * num2))
                              ? (num1 * num1)
                              : (num2 * num2);
                }

                // If level is even and value is odd,
                // find the closest even perfect square
                if (!(lvl & 1) && (x1 & 1)) {

                    int num1 = x1 - 1, num2 = x1 + 1;

                    node->data
                        = (abs(node->data - num1 * num1)
                           < abs(node->data - num2 * num2))
                              ? (num1 * num1)
                              : (num2 * num2);
                }
            }

            // Otherwise, find the find
            // the nearest perfect square
            else {
                if (lvl & 1)
                    node->data
                        = (x1 & 1) ? (x1 * x1) : (x2 * x2);
                else
                    node->data
                        = (x1 & 1) ? (x2 * x2) : (x1 * x1);
            }

            // Print front of queue
            // and remove it from queue
            cout << node->data << " ";
            q.pop();

            // Enqueue left child
            if (node->left != NULL)
                q.push(node->left);

            // Enqueue right child
            if (node->right != NULL)
                q.push(node->right);
        }

        // Increment the level by 1
        lvl++;
        cout << endl;
    }
}

// Utility function to
// create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// Driver Code
int main()
{
    // Binary Tree
    Node* root = newNode(5);
    root->left = newNode(3);
    root->right = newNode(2);
    root->right->left = newNode(16);
    root->right->right = newNode(19);

    LevelOrderTraversal(root);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

// Structure of a
// Binary Tree Node
class Node
{
    int data;
    Node left, right;

    Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}

class GFG{

// Function to replace all nodes
// at even and odd levels with their
// nearest even or odd perfect squares
static void LevelOrderTraversal(Node root)
{

    // Base Case
    if (root == null)
        return;

    // Create an empty queue
    // for level order traversal
    Queue<Node> q = new LinkedList<>();

    // Enqueue root
    q.add(root);

    // Initialize height
    int lvl = 1;

    // Iterate until queue is not empty
    while (q.size() != 0)
    {

        // Store the size
        // of the queue
        int n = q.size();

        // Traverse in range [1, n]
        for(int i = 0; i < n; i++)
        {

            // Store the current node
            Node node = q.peek();

            // Store its square root
            double num = Math.sqrt(node.data);
            int x1 = (int)Math.floor(num);
            int x2 = (int)Math.ceil(num);

            // Check if it is a perfect square
            if (x1 == x2)
            {

                // If level is odd and value is even,
                // find the closest odd perfect square
                if (((lvl & 1) != 0) && !((x1 & 1) != 0))
                {
                    int num1 = x1 - 1, num2 = x1 + 1;

                    node.data = (Math.abs(node.data - num1 * num1) <
                                 Math.abs(node.data - num2 * num2)) ?
                                 (num1 * num1) : (num2 * num2);
                }

                // If level is even and value is odd,
                // find the closest even perfect square
                if (!((lvl & 1) != 0) && ((x1 & 1) != 0))
                {
                    int num1 = x1 - 1, num2 = x1 + 1;

                    node.data = (Math.abs(node.data - num1 * num1) <
                                 Math.abs(node.data - num2 * num2)) ?
                                 (num1 * num1) : (num2 * num2);
                }
            }

            // Otherwise, find the find
            // the nearest perfect square
            else
            {
                if ((lvl & 1) != 0)
                    node.data = (x1 & 1) != 0 ?
                                (x1 * x1) : (x2 * x2);
                else
                    node.data = (x1 & 1) != 0 ?
                                (x2 * x2) : (x1 * x1);
            }

            // Print front of queue
            // and remove it from queue
            System.out.print(node.data + " ");
            q.poll();

            // Enqueue left child
            if (node.left != null)
                q.add(node.left);

            // Enqueue right child
            if (node.right != null)
                q.add(node.right);
        }

        // Increment the level by 1
        lvl++;
        System.out.println();
    }
}

// Driver Code
public static void main (String[] args)
{

    // Binary Tree
    Node root = new Node(5);
    root.left = new Node(3);
    root.right = new Node(2);
    root.right.left = new Node(16);
    root.right.right = new Node(19);

    LevelOrderTraversal(root);
}
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program for the above approach
from collections import deque
from math import sqrt, ceil, floor

# Structure of a
# Binary Tree Node
class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# Function to replace all nodes
# at even and odd levels with their
# nearest even or odd perfect squares
def LevelOrderTraversal(root):

    # Base Case
    if (root == None):
        return

    # Create an empty queue
    # for level order traversal
    q = deque()

    # Enqueue root
    q.append(root)

    # Initialize height
    lvl = 1

    # Iterate until queue is not empty
    while (len(q) > 0):

        # Store the size
        # of the queue
        n = len(q)

        # Traverse in range [1, n]
        for i in range(n):

            # Store the current node
            node = q.popleft()

            # Store its square root
            num = sqrt(node.data)
            x1 = floor(num)
            x2 = ceil(num)

            # Check if it is a perfect square
            if (x1 == x2):

                # If level is odd and value is even,
                # find the closest odd perfect square
                if ((lvl & 1) and not (x1 & 1)):
                    num1, num2 = x1 - 1, x1 + 1
                    node.data = (num1 * num1) if (abs(node.data - num1 * num1) < abs(node.data - num2 * num2)) else (num2 * num2)

                # If level is even and value is odd,
                # find the closest even perfect square
                if (not (lvl & 1) and (x1 & 1)):
                    num1,num2 = x1 - 1, x1 + 1
                    node.data = (num1 * num1) if (abs(node.data - num1 * num1) < abs(node.data - num2 * num2)) else (num2 * num2)

            # Otherwise, find the find
            # the nearest perfect square
            else:
                if (lvl & 1):
                    node.data = (x1 * x1) if (x1 & 1) else (x2 * x2)
                else:
                    node.data = (x2 * x2) if (x1 & 1) else (x1 * x1)

            # Prfront of queue
            # and remove it from queue
            print(node.data, end = " ")

            # Enqueue left child
            if (node.left != None):
                q.append(node.left)

            # Enqueue right child
            if (node.right != None):
                q.append(node.right)

        # Increment the level by 1
        lvl += 1
        print()

# Driver Code
if __name__ == '__main__':

    # Binary Tree
    root = Node(5)
    root.left = Node(3)
    root.right = Node(2)
    root.right.left = Node(16)
    root.right.right = Node(19)

    LevelOrderTraversal(root)

    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

// Structure of a
// Binary Tree Node
public class Node
{
  public int data;
  public Node left, right;

  public Node(int data)
  {
    this.data = data;
    left = right = null;
  }
}

public class GFG
{

  // Function to replace all nodes
  // at even and odd levels with their
  // nearest even or odd perfect squares
  static void LevelOrderTraversal(Node root)
  {

    // Base Case
    if (root == null)
      return;

    // Create an empty queue
    // for level order traversal
    Queue<Node> q = new Queue<Node>();

    // Enqueue root
    q.Enqueue(root);

    // Initialize height
    int lvl = 1;

    // Iterate until queue is not empty
    while (q.Count != 0)
    {

      // Store the size
      // of the queue
      int n = q.Count;

      // Traverse in range [1, n]
      for(int i = 0; i < n; i++)
      {

        // Store the current node
        Node node = q.Peek();

        // Store its square root
        double num = Math.Sqrt(node.data);
        int x1 = (int)Math.Floor(num);
        int x2 = (int)Math.Ceiling(num);

        // Check if it is a perfect square
        if (x1 == x2)
        {

          // If level is odd and value is even,
          // find the closest odd perfect square
          if (((lvl & 1) != 0) && !((x1 & 1) != 0))
          {
            int num1 = x1 - 1, num2 = x1 + 1;

            node.data = (Math.Abs(node.data - num1 * num1) <
                         Math.Abs(node.data - num2 * num2)) ?
              (num1 * num1) : (num2 * num2);
          }

          // If level is even and value is odd,
          // find the closest even perfect square
          if (!((lvl & 1) != 0) && ((x1 & 1) != 0))
          {
            int num1 = x1 - 1, num2 = x1 + 1;

            node.data = (Math.Abs(node.data - num1 * num1) <
                         Math.Abs(node.data - num2 * num2)) ?
              (num1 * num1) : (num2 * num2);
          }
        }

        // Otherwise, find the find
        // the nearest perfect square
        else
        {
          if ((lvl & 1) != 0)
            node.data = (x1 & 1) != 0 ?
            (x1 * x1) : (x2 * x2);
          else
            node.data = (x1 & 1) != 0 ?
            (x2 * x2) : (x1 * x1);
        }

        // Print front of queue
        // and remove it from queue
        Console.Write(node.data + " ");
        q.Dequeue();

        // Enqueue left child
        if (node.left != null)
          q.Enqueue(node.left);

        // Enqueue right child
        if (node.right != null)
          q.Enqueue(node.right);
      }

      // Increment the level by 1
      lvl++;
      Console.WriteLine();
    }
  }

  // Driver Code
  static public void Main ()
  {

    // Binary Tree
    Node root = new Node(5);
    root.left = new Node(3);
    root.right = new Node(2);
    root.right.left = new Node(16);
    root.right.right = new Node(19);

    LevelOrderTraversal(root);
  }
}

// This code is contributed by rag2127

Javascript

<script>

    // JavaScript program for the above approach

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    // Function to replace all nodes
    // at even and odd levels with their
    // nearest even or odd perfect squares
    function LevelOrderTraversal(root)
    {

        // Base Case
        if (root == null)
            return;

        // Create an empty queue
        // for level order traversal
        let q = [];

        // Enqueue root
        q.push(root);

        // Initialize height
        let lvl = 1;

        // Iterate until queue is not empty
        while (q.length != 0)
        {

            // Store the size
            // of the queue
            let n = q.length;

            // Traverse in range [1, n]
            for(let i = 0; i < n; i++)
            {

                // Store the current node
                let node = q[0];

                // Store its square root
                let num = Math.sqrt(node.data);
                let x1 = Math.floor(num);
                let x2 = Math.ceil(num);

                // Check if it is a perfect square
                if (x1 == x2)
                {

                    // If level is odd and value is even,
                    // find the closest odd perfect square
                    if (((lvl & 1) != 0) && !((x1 & 1) != 0))
                    {
                      let num1 = x1 - 1, num2 = x1 + 1;

                      node.data = (Math.abs(node.data - num1 * num1) <
                                   Math.abs(node.data - num2 * num2))?
                                     (num1 * num1) : (num2 * num2);
                    }

                    // If level is even and value is odd,
                    // find the closest even perfect square
                    if (!((lvl & 1) != 0) && ((x1 & 1) != 0))
                    {
                      let num1 = x1 - 1, num2 = x1 + 1;

                      node.data = (Math.abs(node.data - num1 * num1) <
                                  Math.abs(node.data - num2 * num2)) ?
                                     (num1 * num1) : (num2 * num2);
                    }
                }

                // Otherwise, find the find
                // the nearest perfect square
                else
                {
                    if ((lvl & 1) != 0)
                        node.data = (x1 & 1) != 0 ?
                                    (x1 * x1) : (x2 * x2);
                    else
                        node.data = (x1 & 1) != 0 ?
                                    (x2 * x2) : (x1 * x1);
                }

                // Print front of queue
                // and remove it from queue
                document.write(node.data + " ");
                q.shift();

                // Enqueue left child
                if (node.left != null)
                    q.push(node.left);

                // Enqueue right child
                if (node.right != null)
                    q.push(node.right);
            }

            // Increment the level by 1
            lvl++;
            document.write("</br>");
        }
    }

    // Binary Tree
    let root = new Node(5);
    root.left = new Node(3);
    root.right = new Node(2);
    root.right.left = new Node(16);
    root.right.right = new Node(19);

    LevelOrderTraversal(root);

</script>

Output

9 
4 4 
9 25 

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)