Dado un árbol binario, la tarea es imprimir todos los Nodes excepto el más a la derecha en cada nivel del árbol. La raíz se considera en el nivel 0 y el Node más a la derecha de cualquier nivel se considera como un Node en la posición 0.
Ejemplos:
Input:
1
/ \
2 3
/ \ \
4 5 6
/ \
7 8
/ \
9 10
Output:
2
4 5
7
9
Input:
1
/ \
2 3
\ \
4 5
Output:
2
4
Enfoque: para imprimir los Nodes nivel por nivel, use el recorrido de orden de niveles. La idea se basa en el recorrido del orden de nivel de impresión línea por línea . Para eso, recorra los Nodes nivel por nivel y si el Node en la cola de orden de nivel es el último Node, ese Node será el Node más a la derecha y no imprimirá ese Node.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Structure of the tree node
struct Node {
int data;
Node *left, *right;
};
// 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);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
void excluderightmost(Node* root)
{
// Base Case
if (root == NULL)
return;
// Create an empty queue for level
// order traversal
queue<Node*> q;
// Enqueue root
q.push(root);
while (1) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.size();
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node* node = q.front();
// If node is not rightmost print
if (nodeCount != 1)
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
cout << "\n";
}
}
// 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->right->right = newNode(10);
excluderightmost(root);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class Sol {
// Structure of the tree node
static class Node {
int data;
Node left, right;
};
// 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);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
static void excluderightmost(Node root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for level
// order traversal
Queue<Node> q = new LinkedList<Node>();
// Enqueue root
q.add(root);
while (true) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.size();
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node node = q.peek();
// If node is not rightmost print
if (nodeCount != 1)
System.out.print(node.data + " ");
q.remove();
if (node.left != null)
q.add(node.left);
if (node.right != null)
q.add(node.right);
nodeCount--;
}
System.out.println();
}
}
// 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.right.right = newNode(10);
excluderightmost(root);
}
}
Python
# Python implementation of the approach from collections import deque # Structure of the tree node class Node: def __init__(self): self.data = 0 self.left = None self.right = None # Utility method to create a node def newNode(data: int) -> Node: node = Node() node.data = data node.left = None node.right = None return node # Function to print all the nodes # except the rightmost in every level # of the given binary tree # with level order traversal def excluderightmost(root: Node): # Base Case if root is None: return # Create an empty queue for level # order traversal q = deque() # Enqueue root q.append(root) while 1: # nodeCount (queue size) indicates # number of nodes at current level nodeCount = len(q) if nodeCount == 0: break # Dequeue all nodes of current level # and Enqueue all nodes of next level while nodeCount > 0: node = q[0] # If Node is not right most print if nodeCount != 1: print(node.data, end =" ") q.popleft() if node.left is not None: q.append(node.left) if node.right is not None: q.append(node.right) nodeCount -= 1 print() # Driver Code if __name__ == "__main__": root = 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.right.right = newNode(10) excluderightmost(root)
C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG {
// Structure of the tree node
public class Node {
public int data;
public Node left, right;
};
// 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);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
static void excluderightmost(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);
while (true) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.Count;
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node node = q.Peek();
// if Node is not right most print
if (nodeCount != 1)
Console.Write(node.data + " ");
q.Dequeue();
if (node.left != null)
q.Enqueue(node.left);
if (node.right != null)
q.Enqueue(node.right);
nodeCount--;
}
Console.WriteLine();
}
}
// 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.right.right = newNode(10);
excluderightmost(root);
}
}
Javascript
<script>
// JavaScript implementation of the above approach
// Structure of the tree node
class Node {
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
}
};
// Utility method to create a node
function newNode(data)
{
var node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
function excluderightmost(root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for level
// order traversal
var q = [];
// push root
q.push(root);
while (true) {
// nodeCount (queue size) indicates
// number of nodes at current level.
var nodeCount = q.length;
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and push all nodes of next level
while (nodeCount > 0) {
var node = q[0];
// if Node is not right most print
if (nodeCount != 1)
document.write(node.data + " ");
q.shift();
if (node.left != null)
q.push(node.left);
if (node.right != null)
q.push(node.right);
nodeCount--;
}
document.write("<br>");
}
}
// Driver code
var 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.right.right = newNode(10);
excluderightmost(root);
</script>
Producción:
2 4 5 6 8
Publicación traducida automáticamente
Artículo escrito por SHUBHAMSINGH10 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA