Dado un árbol binario que consiste en N Nodes, la tarea es reemplazar todos los Nodes que están presentes en los niveles pares en un árbol binario con su cuadrado perfecto par más cercano y reemplazar los Nodes en los niveles impares con su cuadrado perfecto impar más cercano .
Ejemplos:
Entrada: 5
/ \
3 2
/ \
16 19Salida: 9
/ \
4 4
/ \
9 25Explicación:
Nivel 1: El cuadrado perfecto impar más cercano a 5 es 9.
Nivel 2: El cuadrado perfecto par más cercano a 3 y 2 son 4.
Nivel 3: El cuadrado perfecto impar más cercano a 16 es 9 y a 19 es 25.Entrada: 45
/ \
65 32
/
89Salida: 49
/ \
64 36
/
81Explicación:
Nivel 1: El cuadrado perfecto impar más cercano a 45 es 49.
Nivel 2: El cuadrado perfecto par más cercano a 65 y 32 son 64 y 36 respectivamente.
Nivel 3: El cuadrado perfecto impar más cercano a 89 es 81.
Enfoque: El problema dado se puede resolver usando Level Order Traversal . Siga los pasos a continuación para resolver el problema:
- Inicialice una cola , diga q.
- Empuje la raíz en la cola.
- Bucle mientras la cola no está vacía
- Almacene el Node actual en una variable, digamos temp_node .
- Si el valor del Node actual es un cuadrado perfecto , verifique lo siguiente:
- Si el valor del nivel es impar y el valor del Node actual es par, encuentre el cuadrado perfecto impar más cercano y actualice temp_node→data .
- Si el valor del nivel es par y el valor del Node actual es impar, encuentre el cuadrado perfecto par más cercano y actualice temp_node→data .
- De lo contrario, encuentre el cuadrado perfecto impar más cercano si el valor del nivel es impar o el cuadrado perfecto par más cercano si el valor del nivel es par. Actualice temp_node→data .
- Imprimir temp_node→datos .
- Encola los hijos de temp_node (primero el hijo izquierdo , luego el hijo derecho ) a q , si existe.
- Retire el Node actual de q .
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
9 4 4 9 25
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por rohitsingh07052 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA