Dado un árbol ternario, cree una lista doblemente enlazada a partir de él. Un árbol ternario es como un árbol binario, pero en lugar de tener dos Nodes, tiene tres Nodes, es decir, izquierdo, medio y derecho.
La lista doblemente enlazada debe tener las siguientes propiedades:
- El puntero izquierdo del árbol ternario debería actuar como puntero anterior de la lista doblemente enlazada.
- El puntero del medio del árbol ternario no debe apuntar a nada.
- El puntero derecho del árbol ternario debe actuar como el siguiente puntero de la lista doblemente enlazada.
- Cada Node del árbol ternario se inserta en la lista doblemente enlazada antes de sus subárboles y para cualquier Node, su hijo izquierdo se insertará primero, seguido por el hijo medio y derecho (si lo hay).
Para el ejemplo anterior, la lista enlazada formada para el siguiente árbol debe ser NULL <- 30 <-> 5 <-> 1 <-> 4 <-> 8 <-> 11 <-> 6 <-> 7 <-> 15 <-> 63 <-> 31 <-> 55 <-> 65 -> NULO
Le recomendamos encarecidamente que minimice su navegador y que pruebe esto usted mismo primero.
La idea es atravesar el árbol en orden anticipado similar al recorrido en orden previo del árbol binario. Aquí, cuando visitemos un Node, lo insertaremos en una lista doblemente enlazada, al final, usando un puntero de cola. Que utilizamos para mantener el orden de inserción requerido. Luego llamamos recursivamente al hijo izquierdo, al hijo del medio y al hijo derecho en ese orden.
A continuación se muestra la implementación de esta idea.
C++
// C++ program to create a doubly linked list out
// of given a ternary tree.
#include <bits/stdc++.h>
using namespace std;
/* A ternary tree */
struct Node
{
int data;
struct Node *left, *middle, *right;
};
/* Helper function that allocates a new node with the
given data and assign NULL to left, middle and right
pointers.*/
Node* newNode(int data)
{
Node* node = new Node;
node->data = data;
node->left = node->middle = node->right = NULL;
return node;
}
/* Utility function that constructs doubly linked list
by inserting current node at the end of the doubly
linked list by using a tail pointer */
void push(Node** tail_ref, Node* node)
{
// initialize tail pointer
if (*tail_ref == NULL)
{
*tail_ref = node;
// set left, middle and right child to point
// to NULL
node->left = node->middle = node->right = NULL;
return;
}
// insert node in the end using tail pointer
(*tail_ref)->right = node;
// set prev of node
node->left = (*tail_ref);
// set middle and right child to point to NULL
node->right = node->middle = NULL;
// now tail pointer will point to inserted node
(*tail_ref) = node;
}
/* Create a doubly linked list out of given a ternary tree.
by traversing the tree in preorder fashion. */
void TernaryTreeToList(Node* root, Node** head_ref)
{
// Base case
if (root == NULL)
return;
//create a static tail pointer
static Node* tail = NULL;
// store left, middle and right nodes
// for future calls.
Node* left = root->left;
Node* middle = root->middle;
Node* right = root->right;
// set head of the doubly linked list
// head will be root of the ternary tree
if (*head_ref == NULL)
*head_ref = root;
// push current node in the end of DLL
push(&tail, root);
//recurse for left, middle and right child
TernaryTreeToList(left, head_ref);
TernaryTreeToList(middle, head_ref);
TernaryTreeToList(right, head_ref);
}
// Utility function for printing double linked list.
void printList(Node* head)
{
printf("Created Double Linked list is:\n");
while (head)
{
printf("%d ", head->data);
head = head->right;
}
}
// Driver program to test above functions
int main()
{
// Constructing ternary tree as shown in above figure
Node* root = newNode(30);
root->left = newNode(5);
root->middle = newNode(11);
root->right = newNode(63);
root->left->left = newNode(1);
root->left->middle = newNode(4);
root->left->right = newNode(8);
root->middle->left = newNode(6);
root->middle->middle = newNode(7);
root->middle->right = newNode(15);
root->right->left = newNode(31);
root->right->middle = newNode(55);
root->right->right = newNode(65);
Node* head = NULL;
TernaryTreeToList(root, &head);
printList(head);
return 0;
}
Java
//Java program to create a doubly linked list
// from a given ternary tree.
//Custom node class.
class newNode
{
int data;
newNode left,middle,right;
public newNode(int data)
{
this.data = data;
left = middle = right = null;
}
}
class GFG {
//tail of the linked list.
static newNode tail;
//function to push the node to the tail.
public static void push(newNode node)
{
//to put the node at the end of
// the already existing tail.
tail.right = node;
//to point to the previous node.
node.left = tail;
// middle pointer should point to
// nothing so null. initiate right
// pointer to null.
node.middle = node.right = null;
//update the tail position.
tail = node;
}
/* Create a doubly linked list out of given a ternary tree.
by traversing the tree in preorder fashion. */
public static void ternaryTree(newNode node,newNode head)
{
if(node == null)
return;
newNode left = node.left;
newNode middle = node.middle;
newNode right = node.right;
if(tail != node)
// already root is in the tail so dont push
// the node when it was root.In the first
// case both node and tail have root in them.
push(node);
// First the left child is to be taken.
// Then middle and then right child.
ternaryTree(left,head);
ternaryTree(middle,head);
ternaryTree(right,head);
}
//function to initiate the list process.
public static newNode startTree(newNode root)
{
//Initiate the head and tail with root.
newNode head = root;
tail = root;
ternaryTree(root,head);
//since the head,root are passed
// with reference the changes in
// root will be reflected in head.
return head;
}
// Utility function for printing double linked list.
public static void printList(newNode head)
{
System.out.print("Created Double Linked list is:\n");
while(head != null)
{
System.out.print(head.data + " ");
head = head.right;
}
}
// Driver program to test above functions
public static void main(String args[])
{
// Constructing ternary tree as shown
// in above figure
newNode root = new newNode(30);
root.left = new newNode(5);
root.middle = new newNode(11);
root.right = new newNode(63);
root.left.left = new newNode(1);
root.left.middle = new newNode(4);
root.left.right = new newNode(8);
root.middle.left = new newNode(6);
root.middle.middle = new newNode(7);
root.middle.right = new newNode(15);
root.right.left = new newNode(31);
root.right.middle = new newNode(55);
root.right.right = new newNode(65);
// The function which initiates the list
// process returns the head.
newNode head = startTree(root);
printList(head);
}
}
// This code is contributed by M.V.S.Surya Teja.
Python3
# Python3 program to create a doubly linked
# list out of given a ternary tree.
# Custom node class.
class newNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.middle = None
class GFG:
def __init__(self):
# Tail of the linked list.
self.tail = None
# Function to push the node to the tail.
def push(self, node):
# To put the node at the end of
# the already existing tail.
self.tail.right = node
# To point to the previous node.
node.left = self.tail
# Middle pointer should point to
# nothing so null. initiate right
# pointer to null.
node.middle = node.right = None
# Update the tail position.
self.tail = node
# Create a doubly linked list out of given
# a ternary tree By traversing the tree in
# preorder fashion.
def ternaryTree(self, node, head):
if node == None:
return
left = node.left
middle = node.middle
right = node.right
if self.tail != node:
# Already root is in the tail so dont push
# the node when it was root.In the first
# case both node and tail have root in them.
self.push(node)
# First the left child is to be taken.
# Then middle and then right child.
self.ternaryTree(left, head)
self.ternaryTree(middle, head)
self.ternaryTree(right, head)
def startTree(self, root):
# Initiate the head and tail with root.
head = root
self.tail = root
self.ternaryTree(root, head)
# Since the head,root are passed
# with reference the changes in
# root will be reflected in head.
return head
# Utility function for printing double linked list.
def printList(self, head):
print("Created Double Linked list is:")
while head:
print(head.data, end = " ")
head = head.right
# Driver code
if __name__ == '__main__':
# Constructing ternary tree as shown
# in above figure
root = newNode(30)
root.left = newNode(5)
root.middle = newNode(11)
root.right = newNode(63)
root.left.left = newNode(1)
root.left.middle = newNode(4)
root.left.right = newNode(8)
root.middle.left = newNode(6)
root.middle.middle = newNode(7)
root.middle.right = newNode(15)
root.right.left = newNode(31)
root.right.middle = newNode(55)
root.right.right = newNode(65)
# The function which initiates the list
# process returns the head.
head = None
gfg = GFG()
head = gfg.startTree(root)
gfg.printList(head)
# This code is contributed by Winston Sebastian Pais
C#
// C# program to create a doubly linked
// list from a given ternary tree.
using System;
// Custom node class.
public class newNode
{
public int data;
public newNode left, middle, right;
public newNode(int data)
{
this.data = data;
left = middle = right = null;
}
}
class GFG
{
// tail of the linked list.
public static newNode tail;
// function to push the node to the tail.
public static void push(newNode node)
{
// to put the node at the end of
// the already existing tail.
tail.right = node;
// to point to the previous node.
node.left = tail;
// middle pointer should point to
// nothing so null. initiate right
// pointer to null.
node.middle = node.right = null;
// update the tail position.
tail = node;
}
/* Create a doubly linked list out
of given a ternary tree. by traversing
the tree in preorder fashion. */
public static void ternaryTree(newNode node,
newNode head)
{
if (node == null)
{
return;
}
newNode left = node.left;
newNode middle = node.middle;
newNode right = node.right;
if (tail != node)
{
// already root is in the tail so dont push
// the node when it was root.In the first
// case both node and tail have root in them.
push(node);
}
// First the left child is to be taken.
// Then middle and then right child.
ternaryTree(left, head);
ternaryTree(middle, head);
ternaryTree(right, head);
}
// function to initiate the list process.
public static newNode startTree(newNode root)
{
// Initiate the head and tail with root.
newNode head = root;
tail = root;
ternaryTree(root,head);
// since the head,root are passed
// with reference the changes in
// root will be reflected in head.
return head;
}
// Utility function for printing
// double linked list.
public static void printList(newNode head)
{
Console.Write("Created Double Linked list is:\n");
while (head != null)
{
Console.Write(head.data + " ");
head = head.right;
}
}
// Driver Code
public static void Main(string[] args)
{
// Constructing ternary tree as shown
// in above figure
newNode root = new newNode(30);
root.left = new newNode(5);
root.middle = new newNode(11);
root.right = new newNode(63);
root.left.left = new newNode(1);
root.left.middle = new newNode(4);
root.left.right = new newNode(8);
root.middle.left = new newNode(6);
root.middle.middle = new newNode(7);
root.middle.right = new newNode(15);
root.right.left = new newNode(31);
root.right.middle = new newNode(55);
root.right.right = new newNode(65);
// The function which initiates the list
// process returns the head.
newNode head = startTree(root);
printList(head);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
//javascript program to create a doubly linked list
// from a given ternary tree.
//Custom node class.
class newNode {
constructor(data) {
this.data = data;
this.left = null;
this.middle = null;
this.right = null;
}
}
// tail of the linked list.
var tail;
// function to push the node to the tail.
function push( node) {
// to put the node at the end of
// the already existing tail.
tail.right = node;
// to point to the previous node.
node.left = tail;
// middle pointer should point to
// nothing so null. initiate right
// pointer to null.
node.middle = node.right = null;
// update the tail position.
tail = node;
}
/*
* Create a doubly linked list out of given a ternary tree. by traversing the
* tree in preorder fashion.
*/
function ternaryTree( node, head) {
if (node == null)
return;
var left = node.left;
var middle = node.middle;
var right = node.right;
if (tail != node)
// already root is in the tail so dont push
// the node when it was root.In the first
// case both node and tail have root in them.
push(node);
// First the left child is to be taken.
// Then middle and then right child.
ternaryTree(left, head);
ternaryTree(middle, head);
ternaryTree(right, head);
}
// function to initiate the list process.
function startTree( root) {
// Initiate the head and tail with root.
var head = root;
tail = root;
ternaryTree(root, head);
// since the head,root are passed
// with reference the changes in
// root will be reflected in head.
return head;
}
// Utility function for printing var linked list.
function printList( head) {
document.write("Created Double Linked list is:<br/>");
while (head != null) {
document.write(head.data + " ");
head = head.right;
}
}
// Driver program to test above functions
// Constructing ternary tree as shown
// in above figure
root = new newNode(30);
root.left = new newNode(5);
root.middle = new newNode(11);
root.right = new newNode(63);
root.left.left = new newNode(1);
root.left.middle = new newNode(4);
root.left.right = new newNode(8);
root.middle.left = new newNode(6);
root.middle.middle = new newNode(7);
root.middle.right = new newNode(15);
root.right.left = new newNode(31);
root.right.middle = new newNode(55);
root.right.right = new newNode(65);
// The function which initiates the list
// process returns the head.
head = startTree(root);
printList(head);
// This code contributed by gauravrajput1
</script>
Producción
Created Double Linked list is: 30 5 1 4 8 11 6 7 15 63 31 55 65
Complejidad de tiempo: O (n), ya que estamos usando la recursividad para recorrer n veces. Donde n es el número de Nodes del árbol.
Espacio auxiliar: O(n), ya que estamos usando espacio extra para la lista enlazada.
Este artículo es una contribución de Aditya Goel . Si le gusta GeeksforGeeks y le gustaría contribuir, también puede escribir un artículo y enviarlo 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