Escribe una función para conectar todos los Nodes adyacentes al mismo nivel en un árbol binario.
Ejemplo:
Input Tree
A
/ \
B C
/ \ \
D E F
Output Tree
A--->NULL
/ \
B-->C-->NULL
/ \ \
D-->E-->F-->NULL
Ya hemos discutido el tiempo O (n ^ 2) y el enfoque O en los Nodes de conexión al mismo nivel que el recorrido de morris en el peor de los casos puede ser O (n) y llamarlo para establecer el puntero derecho puede resultar en una complejidad de tiempo O (n ^ 2) .
En esta publicación, hemos discutido el cruce de orden de nivel con marcadores NULL que se necesitan para marcar niveles en el árbol.
Implementación:
C++
// Connect nodes at same level using level order
// traversal.
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* left, *right, *nextRight;
};
// Sets nextRight of all nodes of a tree
void connect(struct Node* root)
{
queue<Node*> q;
q.push(root);
// null marker to represent end of current level
q.push(NULL);
// Do Level order of tree using NULL markers
while (!q.empty()) {
Node *p = q.front();
q.pop();
if (p != NULL) {
// next element in queue represents next
// node at current Level
p->nextRight = q.front();
// push left and right children of current
// node
if (p->left)
q.push(p->left);
if (p->right)
q.push(p->right);
}
// if queue is not empty, push NULL to mark
// nodes at this level are visited
else if (!q.empty())
q.push(NULL);
}
}
/* UTILITY FUNCTIONS */
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* newnode(int data)
{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->data = data;
node->left = node->right = node->nextRight = NULL;
return (node);
}
/* Driver program to test above functions*/
int main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
struct Node* root = newnode(10);
root->left = newnode(8);
root->right = newnode(2);
root->left->left = newnode(3);
root->right->right = newnode(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
printf("Following are populated nextRight pointers in \n"
"the tree (-1 is printed if there is no nextRight) \n");
printf("nextRight of %d is %d \n", root->data,
root->nextRight ? root->nextRight->data : -1);
printf("nextRight of %d is %d \n", root->left->data,
root->left->nextRight ? root->left->nextRight->data : -1);
printf("nextRight of %d is %d \n", root->right->data,
root->right->nextRight ? root->right->nextRight->data : -1);
printf("nextRight of %d is %d \n", root->left->left->data,
root->left->left->nextRight ? root->left->left->nextRight->data : -1);
printf("nextRight of %d is %d \n", root->right->right->data,
root->right->right->nextRight ? root->right->right->nextRight->data : -1);
return 0;
}
Java
// Connect nodes at same level using level order
// traversal.
import java.util.LinkedList;
import java.util.Queue;
public class Connect_node_same_level {
// Node class
static class Node {
int data;
Node left, right, nextRight;
Node(int data){
this.data = data;
left = null;
right = null;
nextRight = null;
}
};
// Sets nextRight of all nodes of a tree
static void connect(Node root)
{
Queue<Node> q = new LinkedList<Node>();
q.add(root);
// null marker to represent end of current level
q.add(null);
// Do Level order of tree using NULL markers
while (!q.isEmpty()) {
Node p = q.peek();
q.remove();
if (p != null) {
// next element in queue represents next
// node at current Level
p.nextRight = q.peek();
// push left and right children of current
// node
if (p.left != null)
q.add(p.left);
if (p.right != null)
q.add(p.right);
}
// if queue is not empty, push NULL to mark
// nodes at this level are visited
else if (!q.isEmpty())
q.add(null);
}
}
/* Driver program to test above functions*/
public static void main(String args[])
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
Node root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
System.out.println("Following are populated nextRight pointers in \n" +
"the tree (-1 is printed if there is no nextRight)");
System.out.println("nextRight of "+ root.data +" is "+
((root.nextRight != null) ? root.nextRight.data : -1));
System.out.println("nextRight of "+ root.left.data+" is "+
((root.left.nextRight != null) ? root.left.nextRight.data : -1));
System.out.println("nextRight of "+ root.right.data+" is "+
((root.right.nextRight != null) ? root.right.nextRight.data : -1));
System.out.println("nextRight of "+ root.left.left.data+" is "+
((root.left.left.nextRight != null) ? root.left.left.nextRight.data : -1));
System.out.println("nextRight of "+ root.right.right.data+" is "+
((root.right.right.nextRight != null) ? root.right.right.nextRight.data : -1));
}
}
// This code is contributed by Sumit Ghosh
Python3
#! /usr/bin/env python3
# connect nodes at same level using level order traversal
import sys
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.nextRight = None
def __str__(self):
return '{}'.format(self.data)
def printLevelByLevel(root):
# print level by level
if root:
node = root
while node:
print('{}'.format(node.data), end=' ')
node = node.nextRight
print()
if root.left:
printLevelByLevel(root.left)
else:
printLevelByLevel(root.right)
def inorder(root):
if root:
inorder(root.left)
print(root.data, end=' ')
inorder(root.right)
def connect(root):
# set nextRight of all nodes of a tree
queue = []
queue.append(root)
# null marker to represent end of current level
queue.append(None)
# do level order of tree using None markers
while queue:
p = queue.pop(0)
if p:
# next element in queue represents
# next node at current level
p.nextRight = queue[0]
# pus left and right children of current node
if p.left:
queue.append(p.left)
if p.right:
queue.append(p.right)
else if queue:
queue.append(None)
def main():
"""Driver program to test above functions.
Constructed binary tree is
10
/ \
8 2
/ \
3 90
"""
root = Node(10)
root.left = Node(8)
root.right = Node(2)
root.left.left = Node(3)
root.right.right = Node(90)
# Populates nextRight pointer in all nodes
connect(root)
# Let us check the values of nextRight pointers
print("Following are populated nextRight pointers in \n"
"the tree (-1 is printed if there is no nextRight) \n")
if(root.nextRight != None):
print("nextRight of %d is %d \n" %(root.data,root.nextRight.data))
else:
print("nextRight of %d is %d \n" %(root.data,-1))
if(root.left.nextRight != None):
print("nextRight of %d is %d \n" %(root.left.data,root.left.nextRight.data))
else:
print("nextRight of %d is %d \n" %(root.left.data,-1))
if(root.right.nextRight != None):
print("nextRight of %d is %d \n" %(root.right.data,root.right.nextRight.data))
else:
print("nextRight of %d is %d \n" %(root.right.data,-1))
if(root.left.left.nextRight != None):
print("nextRight of %d is %d \n" %(root.left.left.data,root.left.left.nextRight.data))
else:
print("nextRight of %d is %d \n" %(root.left.left.data,-1))
if(root.right.right.nextRight != None):
print("nextRight of %d is %d \n" %(root.right.right.data,root.right.right.nextRight.data))
else:
print("nextRight of %d is %d \n" %(root.right.right.data,-1))
print()
if __name__ == "__main__":
main()
# This code is contributed by Ram Basnet
C#
// Connect nodes at same level using level order
// traversal.
using System;
using System.Collections.Generic;
public class Connect_node_same_level
{
// Node class
class Node
{
public int data;
public Node left, right, nextRight;
public Node(int data)
{
this.data = data;
left = null;
right = null;
nextRight = null;
}
};
// Sets nextRight of all nodes of a tree
static void connect(Node root)
{
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
// null marker to represent end of current level
q.Enqueue(null);
// Do Level order of tree using NULL markers
while (q.Count!=0)
{
Node p = q.Peek();
q.Dequeue();
if (p != null)
{
// next element in queue represents next
// node at current Level
p.nextRight = q.Peek();
// push left and right children of current
// node
if (p.left != null)
q.Enqueue(p.left);
if (p.right != null)
q.Enqueue(p.right);
}
// if queue is not empty, push NULL to mark
// nodes at this level are visited
else if (q.Count!=0)
q.Enqueue(null);
}
}
/* Driver code*/
public static void Main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
Node root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
Console.WriteLine("Following are populated nextRight pointers in \n" +
"the tree (-1 is printed if there is no nextRight)");
Console.WriteLine("nextRight of "+ root.data +" is "+
((root.nextRight != null) ? root.nextRight.data : -1));
Console.WriteLine("nextRight of "+ root.left.data+" is "+
((root.left.nextRight != null) ? root.left.nextRight.data : -1));
Console.WriteLine("nextRight of "+ root.right.data+" is "+
((root.right.nextRight != null) ? root.right.nextRight.data : -1));
Console.WriteLine("nextRight of "+ root.left.left.data+" is "+
((root.left.left.nextRight != null) ? root.left.left.nextRight.data : -1));
Console.WriteLine("nextRight of "+ root.right.right.data+" is "+
((root.right.right.nextRight != null) ? root.right.right.nextRight.data : -1));
}
}
/* This code is contributed by Rajput-Ji*/
Javascript
<script>
// Connect nodes at same level using level order traversal.
// A Binary Tree Node
class Node
{
constructor(data, nextRight) {
this.left = null;
this.right = null;
this.data = data;
this.nextRight = nextRight;
}
}
// Sets nextRight of all nodes of a tree
function connect(root)
{
let q = [];
q.push(root);
// null marker to represent end of current level
q.push(null);
// Do Level order of tree using NULL markers
while (q.length > 0) {
let p = q[0];
q.shift();
if (p != null) {
// next element in queue represents next
// node at current Level
p.nextRight = q[0];
// push left and right children of current
// node
if (p.left != null)
q.push(p.left);
if (p.right != null)
q.push(p.right);
}
// if queue is not empty, push NULL to mark
// nodes at this level are visited
else if (q.length > 0)
q.push(null);
}
}
/* Constructed binary tree is
10
/ \
8 2
/ \
3 90
*/
let root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.right.right = new Node(90);
// Populates nextRight pointer in all nodes
connect(root);
// Let us check the values of nextRight pointers
document.write("Following are populated nextRight pointers in " + "</br>" +
"the tree (-1 is printed if there is no nextRight)" + "</br>");
document.write("nextRight of "+ root.data +" is "+
((root.nextRight != null) ? root.nextRight.data : -1) + "</br>");
document.write("nextRight of "+ root.left.data+" is "+
((root.left.nextRight != null) ? root.left.nextRight.data : -1) + "</br>");
document.write("nextRight of "+ root.right.data+" is "+
((root.right.nextRight != null) ? root.right.nextRight.data : -1) + "</br>");
document.write("nextRight of "+ root.left.left.data+" is "+
((root.left.left.nextRight != null) ? root.left.left.nextRight.data : -1) + "</br>");
document.write("nextRight of "+ root.right.right.data+" is "+
((root.right.right.nextRight != null) ? root.right.right.nextRight.data : -1) + "</br>");
// This code is contributed by divyesh072019.
</script>
Producción
Following are populated nextRight pointers in the tree (-1 is printed if there is no nextRight) nextRight of 10 is -1 nextRight of 8 is 2 nextRight of 2 is -1 nextRight of 3 is 90 nextRight of 90 is -1
Complejidad de tiempo: O(n) donde n es el número de Nodes
Espacio auxiliar: O(n) para cola
Implementación alternativa:
También podemos seguir la implementación discutida en Travesía del orden de nivel de impresión línea por línea | conjunto 1 . Seguimos conectando Nodes del mismo nivel haciendo un seguimiento del Node visitado anteriormente del mismo nivel.
Implementación: https://ide.geeksforgeeks.org/gV1Oc2
Gracias a Akilan Sengottaiyan por sugerir esta implementación alternativa.
Este artículo es una contribución de Abhishek Rajput . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA