Dada una lista enlazada individualmente que tiene miembros de datos ordenados en orden ascendente. Construya un árbol de búsqueda binario equilibrado que tenga los mismos miembros de datos que la lista enlazada dada.
Ejemplos:
Input: Linked List 1->2->3
Output: A Balanced BST
2
/ \
1 3
Input: Linked List 1->2->3->4->5->6->7
Output: A Balanced BST
4
/ \
2 6
/ \ / \
1 3 5 7
Input: Linked List 1->2->3->4
Output: A Balanced BST
3
/ \
2 4
/
1
Input: Linked List 1->2->3->4->5->6
Output: A Balanced BST
4
/ \
2 6
/ \ /
1 3 5
Método 1 (Simple) El
siguiente es un algoritmo simple en el que primero encontramos el Node medio de la lista y lo convertimos en la raíz del árbol que se va a construir.
1) Get the Middle of the linked list and make it root.
2) Recursively do same for the left half and right half.
a) Get the middle of the left half and make it left child of the root
created in step 1.
b) Get the middle of right half and make it the right child of the
root created in step 1.
Complejidad de tiempo: O(nLogn) donde n es el número de Nodes en la lista enlazada.
Método 2 (complicado)
El método 1 construye el árbol desde la raíz hasta las hojas. En este método, construimos desde las hojas hasta la raíz. La idea es insertar Nodes en BST en el mismo orden en que aparecen en la Lista enlazada para que el árbol pueda construirse con una complejidad de tiempo O(n). Primero contamos el número de Nodes en la lista enlazada dada. Sea la cuenta n. Después de contar los Nodes, tomamos n/2 Nodes a la izquierda y construimos recursivamente el subárbol izquierdo. Después de construir el subárbol izquierdo, asignamos memoria para la raíz y vinculamos el subárbol izquierdo con la raíz. Finalmente, construimos recursivamente el subárbol derecho y lo vinculamos con la raíz.
Mientras construimos el BST, también seguimos moviendo el puntero del encabezado de la lista al siguiente para que tengamos el puntero apropiado en cada llamada recursiva.
A continuación se muestra la implementación del método 2. El código principal que crea el BST equilibrado está resaltado.
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class LNode
{
public:
int data;
LNode* next;
};
/* A Binary Tree node */
class TNode
{
public:
int data;
TNode* left;
TNode* right;
};
TNode* newNode(int data);
int countLNodes(LNode *head);
TNode* sortedListToBSTRecur(LNode **head_ref, int n);
/* This function counts the number of
nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
TNode* sortedListToBST(LNode *head)
{
/*Count the number of nodes in Linked List */
int n = countLNodes(head);
/* Construct BST */
return sortedListToBSTRecur(&head, n);
}
/* The main function that constructs
balanced BST and returns root of it.
head_ref --> Pointer to pointer to
head node of linked list n --> No.
of nodes in Linked List */
TNode* sortedListToBSTRecur(LNode **head_ref, int n)
{
/* Base Case */
if (n <= 0)
return NULL;
/* Recursively construct the left subtree */
TNode *left = sortedListToBSTRecur(head_ref, n/2);
/* Allocate memory for root, and
link the above constructed left
subtree with root */
TNode *root = newNode((*head_ref)->data);
root->left = left;
/* Change head pointer of Linked List
for parent recursive calls */
*head_ref = (*head_ref)->next;
/* Recursively construct the right
subtree and link it with root
The number of nodes in right subtree
is total nodes - nodes in
left subtree - 1 (for root) which is n-n/2-1*/
root->right = sortedListToBSTRecur(head_ref, n - n / 2 - 1);
return root;
}
/* UTILITY FUNCTIONS */
/* A utility function that returns
count of nodes in a given Linked List */
int countLNodes(LNode *head)
{
int count = 0;
LNode *temp = head;
while(temp)
{
temp = temp->next;
count++;
}
return count;
}
/* Function to insert a node
at the beginning of the linked list */
void push(LNode** head_ref, int new_data)
{
/* allocate node */
LNode* new_node = new LNode();
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(LNode *node)
{
while(node!=NULL)
{
cout << node->data << " ";
node = node->next;
}
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
TNode* newNode(int data)
{
TNode* node = new TNode();
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
/* A utility function to
print preorder traversal of BST */
void preOrder(TNode* node)
{
if (node == NULL)
return;
cout<<node->data<<" ";
preOrder(node->left);
preOrder(node->right);
}
/* Driver code*/
int main()
{
/* Start with the empty list */
LNode* head = NULL;
/* Let us create a sorted linked list to test the functions
Created linked list will be 1->2->3->4->5->6->7 */
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
cout<<"Given Linked List ";
printList(head);
/* Convert List to BST */
TNode *root = sortedListToBST(head);
cout<<"\nPreOrder Traversal of constructed BST ";
preOrder(root);
return 0;
}
// This code is contributed by rathbhupendra
C
#include<stdio.h>
#include<stdlib.h>
/* Link list node */
struct LNode
{
int data;
struct LNode* next;
};
/* A Binary Tree node */
struct TNode
{
int data;
struct TNode* left;
struct TNode* right;
};
struct TNode* newNode(int data);
int countLNodes(struct LNode *head);
struct TNode* sortedListToBSTRecur(struct LNode **head_ref, int n);
/* This function counts the number of nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
struct TNode* sortedListToBST(struct LNode *head)
{
/*Count the number of nodes in Linked List */
int n = countLNodes(head);
/* Construct BST */
return sortedListToBSTRecur(&head, n);
}
/* The main function that constructs balanced BST and returns root of it.
head_ref --> Pointer to pointer to head node of linked list
n --> No. of nodes in Linked List */
struct TNode* sortedListToBSTRecur(struct LNode **head_ref, int n)
{
/* Base Case */
if (n <= 0)
return NULL;
/* Recursively construct the left subtree */
struct TNode *left = sortedListToBSTRecur(head_ref, n/2);
/* Allocate memory for root, and link the above constructed left
subtree with root */
struct TNode *root = newNode((*head_ref)->data);
root->left = left;
/* Change head pointer of Linked List for parent recursive calls */
*head_ref = (*head_ref)->next;
/* Recursively construct the right subtree and link it with root
The number of nodes in right subtree is total nodes - nodes in
left subtree - 1 (for root) which is n-n/2-1*/
root->right = sortedListToBSTRecur(head_ref, n-n/2-1);
return root;
}
/* UTILITY FUNCTIONS */
/* A utility function that returns count of nodes in a given Linked List */
int countLNodes(struct LNode *head)
{
int count = 0;
struct LNode *temp = head;
while(temp)
{
temp = temp->next;
count++;
}
return count;
}
/* Function to insert a node at the beginning of the linked list */
void push(struct LNode** head_ref, int new_data)
{
/* allocate node */
struct LNode* new_node =
(struct LNode*) malloc(sizeof(struct LNode));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(struct LNode *node)
{
while(node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct TNode* newNode(int data)
{
struct TNode* node = (struct TNode*)
malloc(sizeof(struct TNode));
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
/* A utility function to print preorder traversal of BST */
void preOrder(struct TNode* node)
{
if (node == NULL)
return;
printf("%d ", node->data);
preOrder(node->left);
preOrder(node->right);
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct LNode* head = NULL;
/* Let us create a sorted linked list to test the functions
Created linked list will be 1->2->3->4->5->6->7 */
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
printf("\n Given Linked List ");
printList(head);
/* Convert List to BST */
struct TNode *root = sortedListToBST(head);
printf("\n PreOrder Traversal of constructed BST ");
preOrder(root);
return 0;
}
Java
class LinkedList {
/* head node of link list */
static LNode head;
/* Link list Node */
class LNode
{
int data;
LNode next, prev;
LNode(int d)
{
data = d;
next = prev = null;
}
}
/* A Binary Tree Node */
class TNode
{
int data;
TNode left, right;
TNode(int d)
{
data = d;
left = right = null;
}
}
/* This function counts the number of nodes in Linked List
and then calls sortedListToBSTRecur() to construct BST */
TNode sortedListToBST()
{
/*Count the number of nodes in Linked List */
int n = countNodes(head);
/* Construct BST */
return sortedListToBSTRecur(n);
}
/* The main function that constructs balanced BST and
returns root of it.
n --> No. of nodes in the Doubly Linked List */
TNode sortedListToBSTRecur(int n)
{
/* Base Case */
if (n <= 0)
return null;
/* Recursively construct the left subtree */
TNode left = sortedListToBSTRecur(n / 2);
/* head_ref now refers to middle node,
make middle node as root of BST*/
TNode root = new TNode(head.data);
// Set pointer to left subtree
root.left = left;
/* Change head pointer of Linked List for parent
recursive calls */
head = head.next;
/* Recursively construct the right subtree and link it
with root. The number of nodes in right subtree is
total nodes - nodes in left subtree - 1 (for root) */
root.right = sortedListToBSTRecur(n - n / 2 - 1);
return root;
}
/* UTILITY FUNCTIONS */
/* A utility function that returns count of nodes in a
given Linked List */
int countNodes(LNode head)
{
int count = 0;
LNode temp = head;
while (temp != null)
{
temp = temp.next;
count++;
}
return count;
}
/* Function to insert a node at the beginning of
the Doubly Linked List */
void push(int new_data)
{
/* allocate node */
LNode new_node = new LNode(new_data);
/* since we are adding at the beginning,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = head;
/* change prev of head node to new node */
if (head != null)
head.prev = new_node;
/* move the head to point to the new node */
head = new_node;
}
/* Function to print nodes in a given linked list */
void printList(LNode node)
{
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
}
/* A utility function to print preorder traversal of BST */
void preOrder(TNode node)
{
if (node == null)
return;
System.out.print(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
/* Driver program to test above functions */
public static void main(String[] args) {
LinkedList llist = new LinkedList();
/* Let us create a sorted linked list to test the functions
Created linked list will be 7->6->5->4->3->2->1 */
llist.push(7);
llist.push(6);
llist.push(5);
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
System.out.println("Given Linked List ");
llist.printList(head);
/* Convert List to BST */
TNode root = llist.sortedListToBST();
System.out.println("");
System.out.println("Pre-Order Traversal of constructed BST ");
llist.preOrder(root);
}
}
// This code has been contributed by Mayank Jaiswal(mayank_24)
Python3
# Python3 implementation of above approach
# Link list node
class LNode :
def __init__(self):
self.data = None
self.next = None
# A Binary Tree node
class TNode :
def __init__(self):
self.data = None
self.left = None
self.right = None
head = None
# This function counts the number of
# nodes in Linked List and then calls
# sortedListToBSTRecur() to construct BST
def sortedListToBST():
global head
# Count the number of nodes in Linked List
n = countLNodes(head)
# Construct BST
return sortedListToBSTRecur(n)
# The main function that constructs
# balanced BST and returns root of it.
# head -. Pointer to pointer to
# head node of linked list n -. No.
# of nodes in Linked List
def sortedListToBSTRecur( n) :
global head
# Base Case
if (n <= 0) :
return None
# Recursively construct the left subtree
left = sortedListToBSTRecur( int(n/2))
# Allocate memory for root, and
# link the above constructed left
# subtree with root
root = newNode((head).data)
root.left = left
# Change head pointer of Linked List
# for parent recursive calls
head = (head).next
# Recursively construct the right
# subtree and link it with root
# The number of nodes in right subtree
# is total nodes - nodes in
# left subtree - 1 (for root) which is n-n/2-1
root.right = sortedListToBSTRecur( n - int(n/2) - 1)
return root
# UTILITY FUNCTIONS
# A utility function that returns
# count of nodes in a given Linked List
def countLNodes(head) :
count = 0
temp = head
while(temp != None):
temp = temp.next
count = count + 1
return count
# Function to insert a node
#at the beginning of the linked list
def push(head, new_data) :
# allocate node
new_node = LNode()
# put in the data
new_node.data = new_data
# link the old list off the new node
new_node.next = (head)
# move the head to point to the new node
(head) = new_node
return head
# Function to print nodes in a given linked list
def printList(node):
while(node != None):
print( node.data ,end= " ")
node = node.next
# Helper function that allocates a new node with the
# given data and None left and right pointers.
def newNode(data) :
node = TNode()
node.data = data
node.left = None
node.right = None
return node
# A utility function to
# print preorder traversal of BST
def preOrder( node) :
if (node == None) :
return
print(node.data, end = " " )
preOrder(node.left)
preOrder(node.right)
# Driver code
# Start with the empty list
head = None
# Let us create a sorted linked list to test the functions
# Created linked list will be 1.2.3.4.5.6.7
head = push(head, 7)
head = push(head, 6)
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
print("Given Linked List " )
printList(head)
# Convert List to BST
root = sortedListToBST()
print("\nPreOrder Traversal of constructed BST ")
preOrder(root)
# This code is contributed by Arnab Kundu
C#
// C# implementation of above approach
using System;
public class LinkedList
{
/* head node of link list */
static LNode head;
/* Link list Node */
class LNode
{
public int data;
public LNode next, prev;
public LNode(int d)
{
data = d;
next = prev = null;
}
}
/* A Binary Tree Node */
class TNode
{
public int data;
public TNode left, right;
public TNode(int d)
{
data = d;
left = right = null;
}
}
/* This function counts the number
of nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
TNode sortedListToBST()
{
/*Count the number of nodes in Linked List */
int n = countNodes(head);
/* Construct BST */
return sortedListToBSTRecur(n);
}
/* The main function that constructs
balanced BST and returns root of it.
n --> No. of nodes in the Doubly Linked List */
TNode sortedListToBSTRecur(int n)
{
/* Base Case */
if (n <= 0)
return null;
/* Recursively construct the left subtree */
TNode left = sortedListToBSTRecur(n / 2);
/* head_ref now refers to middle node,
make middle node as root of BST*/
TNode root = new TNode(head.data);
// Set pointer to left subtree
root.left = left;
/* Change head pointer of Linked List
for parent recursive calls */
head = head.next;
/* Recursively construct the
right subtree and link it
with root. The number of
nodes in right subtree is
total nodes - nodes in left
subtree - 1 (for root) */
root.right = sortedListToBSTRecur(n - n / 2 - 1);
return root;
}
/* UTILITY FUNCTIONS */
/* A utility function that returns count
of nodes in a given Linked List */
int countNodes(LNode head)
{
int count = 0;
LNode temp = head;
while (temp != null)
{
temp = temp.next;
count++;
}
return count;
}
/* Function to insert a node at the beginning of
the Doubly Linked List */
void push(int new_data)
{
/* allocate node */
LNode new_node = new LNode(new_data);
/* since we are adding at the beginning,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = head;
/* change prev of head node to new node */
if (head != null)
head.prev = new_node;
/* move the head to point to the new node */
head = new_node;
}
/* Function to print nodes in a given linked list */
void printList(LNode node)
{
while (node != null)
{
Console.Write(node.data + " ");
node = node.next;
}
}
/* A utility function to print
preorder traversal of BST */
void preOrder(TNode node)
{
if (node == null)
return;
Console.Write(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
/* Driver code */
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
/* Let us create a sorted
linked list to test the functions
Created linked list will be
7->6->5->4->3->2->1 */
llist.push(7);
llist.push(6);
llist.push(5);
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
Console.WriteLine("Given Linked List ");
llist.printList(head);
/* Convert List to BST */
TNode root = llist.sortedListToBST();
Console.WriteLine("");
Console.WriteLine("Pre-Order Traversal of constructed BST ");
llist.preOrder(root);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript implementation of above approach
/* head node of link list */
var head = null;
/* Link list Node */
class LNode
{
constructor(d)
{
this.data = d;
this.next = null;
this.prev = null;
}
}
/* A Binary Tree Node */
class TNode
{
constructor(d)
{
this.data = d;
this.left = null;
this.right = null;
}
}
/* This function counts the number
of nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
function sortedListToBST()
{
/*Count the number of nodes in Linked List */
var n = countNodes(head);
/* Construct BST */
return sortedListToBSTRecur(n);
}
/* The main function that constructs
balanced BST and returns root of it.
n --> No. of nodes in the Doubly Linked List */
function sortedListToBSTRecur(n)
{
/* Base Case */
if (n <= 0)
return null;
/* Recursively construct the left subtree */
var left = sortedListToBSTRecur(parseInt(n / 2));
/* head_ref now refers to middle node,
make middle node as root of BST*/
var root = new TNode(head.data);
// Set pointer to left subtree
root.left = left;
/* Change head pointer of Linked List
for parent recursive calls */
head = head.next;
/* Recursively construct the
right subtree and link it
with root. The number of
nodes in right subtree is
total nodes - nodes in left
subtree - 1 (for root) */
root.right = sortedListToBSTRecur(n - parseInt(n / 2) - 1);
return root;
}
/* UTILITY FUNCTIONS */
/* A utility function that returns count
of nodes in a given Linked List */
function countNodes(head)
{
var count = 0;
var temp = head;
while (temp != null)
{
temp = temp.next;
count++;
}
return count;
}
/* Function to insert a node at the beginning of
the Doubly Linked List */
function push( new_data)
{
/* allocate node */
var new_node = new LNode(new_data);
/* since we are adding at the beginning,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = head;
/* change prev of head node to new node */
if (head != null)
head.prev = new_node;
/* move the head to point to the new node */
head = new_node;
}
/* Function to print nodes in a given linked list */
function printList( node)
{
while (node != null)
{
document.write(node.data + " ");
node = node.next;
}
}
/* A utility function to print
preorder traversal of BST */
function preOrder(node)
{
if (node == null)
return;
document.write(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
/* Driver code */
/* Let us create a sorted
linked list to test the functions
Created linked list will be
7->6->5->4->3->2->1 */
push(7);
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
document.write("Given Linked List ");
printList(head);
/* Convert List to BST */
var root = sortedListToBST();
document.write("<br>");
document.write("Pre-Order Traversal of constructed BST ");
preOrder(root);
</script>
Producción:
Given Linked List 1 2 3 4 5 6 7 PreOrder Traversal of constructed BST 4 2 1 3 6 5 7
Complejidad de tiempo: O(n)
Espacio auxiliar : O (n) para la pila de llamadas desde que se usa la recursividad
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA