Dado un árbol binario (BT), conviértalo en una lista doblemente enlazada (DLL). Los punteros izquierdo y derecho en los Nodes se utilizarán como punteros anterior y siguiente, respectivamente, en la DLL convertida. El orden de los Nodes en DLL debe ser el mismo que en Inorder para el árbol binario dado. El primer Node del recorrido Inorder debe ser el Node principal de la DLL.
Ejemplos:
Entrada:
1
/ \
3 2
Salida:
Orden real: 3 1 2
Orden inverso: 2 1 3
Explicación: El encabezado de la lista enlazada será 3 y el último elemento será 2.
DLL será 3 <=> 1 <= > 2Entrada:
10
/ \
20 30
/ \
40 60
Salida:
Orden real: 40 20 60 10 30
Orden inverso: 30 10 60 20 40
A continuación se presentan varios enfoques que se han discutido anteriormente:
- Convertir un árbol binario dado en una lista doblemente enlazada | Serie 1
- Convertir un árbol binario dado en una lista doblemente enlazada | conjunto 2
- Convertir un árbol binario dado en una lista doblemente enlazada | conjunto 3
- Convertir un árbol binario dado en una lista doblemente enlazada | conjunto 4
Acercarse:
Los enfoques anteriores utilizan la recursividad o la pila para obtener el Inorder Traversal. Este enfoque se basa en Morris Traversal para encontrar Inorder Traversal , que es iterativo y tiene una complejidad espacial de O(1).
La idea es que primero crearemos una lista con un solo enlace mientras hacemos Morris Traversal y luego la convertiremos en una lista con doble enlace colocando el puntero izquierdo de cada Node en el Node anterior en orden.
Siga los pasos mencionados a continuación para implementar la idea:
- Realice Morris Traversal para atravesar el árbol en orden en tiempo lineal y cree la lista enlazada individualmente.
- Ahora recorra la lista enlazada individualmente:
- Cree un enlace entre el Node actual y el predecesor en orden.
- Actualice el Node actual y anterior (predecesor) según corresponda en cada paso y pase al siguiente Node.
- La lista doblemente enlazada generada es la requerida.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to implement the idea
#include <bits/stdc++.h>
using namespace std;
// Structure for tree and linked list
struct Node {
int data;
Node *left, *right;
};
// Utility function for allocating node for
// Binary Tree.
Node* newNode(int data)
{
Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return node;
}
Node* BToDLL(Node* root)
{
Node* curr = root;
// Store head & tail of the linked list
// created so far
Node *head = NULL, *tail = NULL;
// Morris Traversal
while (curr) {
if (curr->left == NULL) {
// If it is to be the first node
// of the desired Linked list
if (head == NULL) {
head = curr;
tail = curr;
}
else {
// Append it to the tail of the
// linked list we have created
// so far & set it as new tail
tail->right = curr;
tail = tail->right;
}
curr = curr->right;
}
else {
// Inorder predecessor
Node* pred = curr->left;
while (pred->right != NULL
&& pred->right != curr) {
pred = pred->right;
}
if (pred->right == NULL) {
pred->right = curr;
curr = curr->left;
}
else {
// Append it to the tail of
// the linked list
// we have created so far & set it
// as new tail
// Note we don't have to unlink
// predecessor
tail->right = curr;
tail = tail->right;
curr = curr->right;
}
}
}
curr = head;
Node* prev = NULL;
// Converting singly linked list to
// doubly linked list
while (curr) {
curr->left = prev;
prev = curr;
curr = curr->right;
}
return head;
}
// Utility function for printing
// double linked list.
void printList(Node* head)
{
printf("Actual order: ");
while (head) {
printf("%d ", head->data);
head = head->right;
}
}
// Utility function for printing
// double linked list in Reverse Order.
void printReverseList(Node* tail)
{
printf("\nReverse Order: ");
while (tail) {
printf("%d ", tail->data);
tail = tail->left;
}
}
// Driver code
int main()
{
/* Constructing below tree
10
/ \
20 30
/ \
40 60 */
Node* root = newNode(10);
root->left = newNode(20);
root->right = newNode(30);
root->left->left = newNode(40);
root->left->right = newNode(60);
// Function call
Node* head = BToDLL(root);
printList(head);
Node* tail = head;
// Finding Tail of DLL
while (tail && tail->right) {
tail = tail->right;
}
printReverseList(tail);
return 0;
}
Java
// Java code to implement the above idea
class GFG {
// structure of tree and linked list.
class Node {
int data;
Node left, right;
}
// Utility function for allocating node for binary tree.
public Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return node;
}
public Node BToDLL(Node root)
{
Node curr = root;
// Store head & tail of the linked list created so
// far
Node head = null, tail = null;
// Morris Traversal
while (curr != null) {
if (curr.left == null) {
// If it is to be the first node of the
// desired Linked list.
if (head == null) {
head = curr;
tail = curr;
}
else {
// Append it to the tail of the linked
// list we have created so far & set it
// as new tail
tail.right = curr;
tail = tail.right;
}
curr = curr.right;
}
else {
// Inorder predecessor
Node pred = curr.left;
while (pred.right != null
&& pred.right != curr) {
pred = pred.right;
}
if (pred.right == null) {
pred.right = curr;
curr = curr.left;
}
else {
// Append it to the tail of the linked
// list we have created so far & set it
// as new tail. Note we don't have to
// unlink predecessor
tail.right = curr;
tail = tail.right;
curr = curr.right;
}
}
}
curr = head;
Node prev = null;
// Converting singly linked list to doubly linked
// list.
while (curr != null) {
curr.left = prev;
prev = curr;
curr = curr.right;
}
return head;
}
// Utility function for printing doubly linked list
public void printList(Node head)
{
System.out.print("Actual order: ");
while (head != null) {
System.out.print(head.data + " ");
head = head.right;
}
}
// Utility function for printing doubly linked list in
// reverse order.
public void printReverseList(Node tail)
{
System.out.print("\nReverse Order: ");
while (tail != null) {
System.out.print(tail.data + " ");
tail = tail.left;
}
}
public static void main(String[] args)
{
GFG list = new GFG();
/* Constructing below tree
10
/ \
20 30
/ \
40 60 */
Node root = list.newNode(10);
root.left = list.newNode(20);
root.right = list.newNode(30);
root.left.left = list.newNode(40);
root.left.right = list.newNode(60);
// Function call
Node head = list.BToDLL(root);
list.printList(head);
Node tail = head;
// Finding tail of DLL
while (tail != null && tail.right != null) {
tail = tail.right;
}
list.printReverseList(tail);
}
}
// This code is contributed by lokesh (lokeshmvs21).
C#
// C# code to implement the above idea
using System;
public class GFG {
// structure of tree and linked list.
public class Node {
public int data;
public Node left;
public Node right;
}
// Utility function for allocating node for binary tree.
public Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return node;
}
public Node BToDLL(Node root)
{
Node curr = root;
// Store head & tail of the linked list created so
// far
Node head = null, tail = null;
// Morris Traversal
while (curr != null) {
if (curr.left == null) {
// If it is to be the first node of the
// desired Linked list.
if (head == null) {
head = curr;
tail = curr;
}
else {
// Append it to the tail of the linked
// list we have created so far & set it
// as new tail
tail.right = curr;
tail = tail.right;
}
curr = curr.right;
}
else {
// Inorder predecessor
Node pred = curr.left;
while (pred.right != null
&& pred.right != curr) {
pred = pred.right;
}
if (pred.right == null) {
pred.right = curr;
curr = curr.left;
}
else {
// Append it to the tail of the linked
// list we have created so far & set it
// as new tail. Note we don't have to
// unlink predecessor
tail.right = curr;
tail = tail.right;
curr = curr.right;
}
}
}
curr = head;
Node prev = null;
// Converting singly linked list to doubly linked
// list.
while (curr != null) {
curr.left = prev;
prev = curr;
curr = curr.right;
}
return head;
}
// Utility function for printing doubly linked list
public void printList(Node head)
{
Console.Write("Actual order: ");
while (head != null) {
Console.Write(head.data + " ");
head = head.right;
}
}
// Utility function for printing doubly linked list in
// reverse order.
public void printReverseList(Node tail)
{
Console.Write("\nReverse Order: ");
while (tail != null) {
Console.Write(tail.data + " ");
tail = tail.left;
}
}
static public void Main()
{
// Code
GFG list = new GFG();
/* Constructing below tree
10
/ \
20 30
/ \
40 60 */
Node root = list.newNode(10);
root.left = list.newNode(20);
root.right = list.newNode(30);
root.left.left = list.newNode(40);
root.left.right = list.newNode(60);
// Function call
Node head = list.BToDLL(root);
list.printList(head);
Node tail = head;
// Finding tail of DLL
while (tail != null && tail.right != null) {
tail = tail.right;
}
list.printReverseList(tail);
}
}
// This code is contributed by lokesh (lokeshmvs21).
Producción
Actual order: 40 20 60 10 30 Reverse Order: 30 10 60 20 40
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por chadokarvijay y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA