Dado un árbol binario, la tarea es imprimir los Nodes colocados impares de niveles impares 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: 3 9
Input:
2
/ \
4 15
/ /
45 17
Output: 15
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 printOddLevelOddNodes(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);
printOddLevelOddNodes(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 printOddLevelOddNodes(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);
printOddLevelOddNodes(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 printOddLevelOddNodes(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 not 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) printOddLevelOddNodes(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 printOddLevelOddNodes(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);
printOddLevelOddNodes(root);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript implementation of the 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 printOddLevelOddNodes(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);
printOddLevelOddNodes(root);
</script>
Producción:
3 9
Complejidad de tiempo: O(n), donde es el número de Nodes
Espacio Auxiliar: O(log n)
Publicación traducida automáticamente
Artículo escrito por SHUBHAMSINGH10 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA