Imprima Nodes colocados impares de niveles pares en orden de nivel del árbol binario dado

Dado un árbol binario, la tarea es imprimir los Nodes colocados impares de niveles pares en el recorrido del orden de niveles del árbol. La raíz se considera en el nivel 0 y el Node más a la izquierda de cualquier nivel se considera como un Node en la posición 0 .
Ejemplo: 
 

Input:
           1
         /    \
        2       3
      / \      /  \
     4   5    6    7
        /  \     
       8    9
      /      \
     10       11
Output: 5 7 11

Input:
      2
    /   \
   4     15
  /     /
 45   17
Output: 17

Requisito previo: elementos colocados uniformemente en un nivel uniforme. 
Enfoque: para imprimir Nodes nivel por nivel, use el recorrido de orden de nivel. La idea se basa en el recorrido del orden de nivel de impresión línea por línea . Para eso, atraviese los Nodes nivel por nivel y cambie la bandera de nivel impar después de cada nivel. Del mismo modo, marque el segundo Node en cada nivel como posición impar y cámbielo cada vez que se procese el siguiente Node.
A continuación se muestra la implementación del enfoque anterior:
 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    Node *left, *right;
};
 
// Iterative method to do level order
// traversal line by line
void printEvenLevelOddNodes(Node* root)
{
    // Base Case
    if (root == NULL)
        return;
 
    // Create an empty queue for level
    // order traversal
    queue<Node*> q;
 
    // Enqueue root and initialize level as even
    q.push(root);
    bool evenLevel = true;
 
    while (1) {
 
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Mark 1st node as even positioned
        bool evenNodePosition = true;
 
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0) {
            Node* node = q.front();
 
            // Print only even positioned
            // nodes of even levels
            if (evenLevel && !evenNodePosition)
                cout << node->data << " ";
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
 
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
 
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
 
// Utility method to create a node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->right->left = newNode(8);
    root->left->right->right = newNode(9);
    root->left->right->left->left = newNode(10);
    root->left->right->right->right = newNode(11);
 
    printEvenLevelOddNodes(root);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
class GFG
{
 
static class Node
{
    int data;
    Node left, right;
};
 
// Iterative method to do level order
// traversal line by line
static void printEvenLevelOddNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
 
    // Create an empty queue for level
    // order traversal
    Queue<Node> q = new LinkedList<>();
 
    // Enqueue root and initialize level as even
    q.add(root);
    boolean evenLevel = true;
 
    while (true)
    {
 
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Mark 1st node as even positioned
        boolean evenNodePosition = true;
 
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0)
        {
            Node node = q.peek();
 
            // Print only even positioned
            // nodes of even levels
            if (evenLevel && !evenNodePosition)
                System.out.print(node.data + " ");
 
            q.remove();
            if (node.left != null)
                q.add(node.left);
            if (node.right != null)
                q.add(node.right);
            nodeCount--;
 
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
 
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
 
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Driver code
public static void main(String[] args)
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.right.left = newNode(8);
    root.left.right.right = newNode(9);
    root.left.right.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
 
    printEvenLevelOddNodes(root);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python implementation of the approach
 
# Utility method to create a node
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# Iterative method to do level order
# traversal line by line
def printEvenLevelOddNodes(root):
    # Base Case
    if (root == None):
        return
     
    # Create an empty queue for level
    # order traversal
    q =[]
     
    # Enqueue root and initialize level as even
    q.append(root)
    evenLevel = True
     
    while (1):
         
        # nodeCount (queue size) indicates
        # number of nodes in the current level
        nodeCount = len(q)
        if (nodeCount == 0):
            break
         
        # Mark 1st node as even positioned
        evenNodePosition = True
         
        # Dequeue all the nodes of current level
        # and Enqueue all the nodes of next level
        while (nodeCount > 0):
            node = q[0]
            # Pronly even positioned
            # nodes of even levels
            if evenLevel and not evenNodePosition:
                print(node.data, end =" ")
            q.pop(0)
            if (node.left != None):
                q.append(node.left)
            if (node.right != None):
                q.append(node.right)
            nodeCount-= 1
             
            # Switch the even position flag
            evenNodePosition = not evenNodePosition
         
        # Switch the even level flag
        evenLevel = not evenLevel
     
 
 
# Driver code
if __name__ == '__main__':
     
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.left.right.left = newNode(8)
    root.left.right.right = newNode(9)
    root.left.right.left.left = newNode(10)
    root.left.right.right.right = newNode(11)
 
    printEvenLevelOddNodes(root)

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
     
class GFG
{
public class Node
{
    public int data;
    public Node left, right;
};
 
// Iterative method to do level order
// traversal line by line
static void printEvenLevelOddNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
 
    // Create an empty queue for level
    // order traversal
    Queue<Node> q = new Queue<Node>();
 
    // Enqueue root and initialize level as even
    q.Enqueue(root);
    bool evenLevel = true;
 
    while (true)
    {
 
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.Count;
        if (nodeCount == 0)
            break;
 
        // Mark 1st node as even positioned
        bool evenNodePosition = true;
 
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0)
        {
            Node node = q.Peek();
 
            // Print only even positioned
            // nodes of even levels
            if (evenLevel && !evenNodePosition)
                Console.Write(node.data + " ");
 
            q.Dequeue();
            if (node.left != null)
                q.Enqueue(node.left);
            if (node.right != null)
                q.Enqueue(node.right);
            nodeCount--;
 
            // Switch the even position flag
            evenNodePosition = !evenNodePosition;
        }
 
        // Switch the even level flag
        evenLevel = !evenLevel;
    }
}
 
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Driver code
public static void Main(String[] args)
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.right.left = newNode(8);
    root.left.right.right = newNode(9);
    root.left.right.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
 
    printEvenLevelOddNodes(root);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
    // JavaScript implementation of the above approach
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Iterative method to do level order
    // traversal line by line
    function printEvenLevelOddNodes(root)
    {
        // Base Case
        if (root == null)
            return;
 
        // Create an empty queue for level
        // order traversal
        let q = [];
 
        // Enqueue root and initialize level as even
        q.push(root);
        let evenLevel = true;
 
        while (true)
        {
 
            // nodeCount (queue size) indicates
            // number of nodes in the current level
            let nodeCount = q.length;
            if (nodeCount == 0)
                break;
 
            // Mark 1st node as even positioned
            let evenNodePosition = true;
 
            // Dequeue all the nodes of current level
            // and Enqueue all the nodes of next level
            while (nodeCount > 0)
            {
                let node = q[0];
 
                // Print only even positioned
                // nodes of even levels
                if (evenLevel && !evenNodePosition)
                    document.write(node.data + " ");
 
                q.shift();
                if (node.left != null)
                    q.push(node.left);
                if (node.right != null)
                    q.push(node.right);
                nodeCount--;
 
                // Switch the even position flag
                evenNodePosition = !evenNodePosition;
            }
 
            // Switch the even level flag
            evenLevel = !evenLevel;
        }
    }
 
    // Utility method to create a node
    function newNode(data)
    {
        let node = new Node(data);
        return (node);
    }
     
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.right.left = newNode(8);
    root.left.right.right = newNode(9);
    root.left.right.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
   
    printEvenLevelOddNodes(root);
 
</script>
Producción: 

5 7 11

 

Publicación traducida automáticamente

Artículo escrito por SHUBHAMSINGH10 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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