Dado un árbol de búsqueda binario equilibrado (BST), escriba una función isTripletPresent() que devuelva verdadero si hay un triplete en BST dado con suma igual a 0, de lo contrario, devuelve falso. La complejidad de tiempo esperada es O (n ^ 2) y solo se puede usar O (Logn) espacio adicional. Puede modificar el árbol de búsqueda binaria dado. Tenga en cuenta que la altura de un BST equilibrado siempre es O (Inicio de sesión).
Por ejemplo, isTripletPresent() debería devolver verdadero para el siguiente BST porque hay un triplete con suma 0, el triplete es {-13, 6, 7}.
La solución de fuerza bruta es considerar cada triplete en BST y verificar si la suma suma cero. La complejidad temporal de esta solución será O(n^3).
Una mejor solución es crear una array auxiliar y almacenar el recorrido en orden de BST en la array. La array se ordenará como el recorrido en orden de BST siempre produce datos ordenados. Una vez que tenemos el recorrido Inorder, podemos usar el método 2 de esta publicación para encontrar el triplete con suma igual a 0. Esta solución funciona en tiempo O(n^2), pero requiere espacio auxiliar O(n).
La siguiente es la solución que funciona en tiempo O (n ^ 2) y usa espacio adicional O (Logn) :
- Convertir BST dado a lista doblemente enlazada (DLL)
- Ahora itere a través de cada Node de DLL y si la clave del Node es negativa, busque un par en DLL con una suma igual a la clave del Node actual multiplicada por -1. Para encontrar el par, podemos usar el enfoque utilizado en hasArrayTwoCandidates() en el método 1 de esta publicación.
Implementación:
C++
// A C++ program to check if there
// is a triplet with sum equal to 0 in
// a given BST
#include <bits/stdc++.h>
using namespace std;
// A BST node has key, and left and right pointers
class node
{
public:
int key;
node *left;
node *right;
};
// A function to convert given BST to Doubly
// Linked List. left pointer is used
// as previous pointer and right pointer
// is used as next pointer. The function
// sets *head to point to first and *tail
// to point to last node of converted DLL
void convertBSTtoDLL(node* root, node** head, node** tail)
{
// Base case
if (root == NULL)
return;
// First convert the left subtree
if (root->left)
convertBSTtoDLL(root->left, head, tail);
// Then change left of current root
// as last node of left subtree
root->left = *tail;
// If tail is not NULL, then set right
// of tail as root, else current
// node is head
if (*tail)
(*tail)->right = root;
else
*head = root;
// Update tail
*tail = root;
// Finally, convert right subtree
if (root->right)
convertBSTtoDLL(root->right, head, tail);
}
// This function returns true if there
// is pair in DLL with sum equal to given
// sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
bool isPresentInDLL(node* head, node* tail, int sum)
{
while (head != tail)
{
int curr = head->key + tail->key;
if (curr == sum)
return true;
else if (curr > sum)
tail = tail->left;
else
head = head->right;
}
return false;
}
// The main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
bool isTripletPresent(node *root)
{
// Check if the given BST is empty
if (root == NULL)
return false;
// Convert given BST to doubly linked list. head and tail store the
// pointers to first and last nodes in DLLL
node* head = NULL;
node* tail = NULL;
convertBSTtoDLL(root, &head, &tail);
// Now iterate through every node and
// find if there is a pair with sum
// equal to -1 * head->key where head is current node
while ((head->right != tail) && (head->key < 0))
{
// If there is a pair with sum
// equal to -1*head->key, then return
// true else move forward
if (isPresentInDLL(head->right, tail, -1*head->key))
return true;
else
head = head->right;
}
// If we reach here, then
// there was no 0 sum triplet
return false;
}
// A utility function to create
// a new BST node with key as given num
node* newNode(int num)
{
node* temp = new node();
temp->key = num;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to insert a given key to BST
node* insert(node* root, int key)
{
if (root == NULL)
return newNode(key);
if (root->key > key)
root->left = insert(root->left, key);
else
root->right = insert(root->right, key);
return root;
}
// Driver code
int main()
{
node* root = NULL;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root))
cout << "Present";
else
cout << "Not Present";
return 0;
}
// This code is contributed by rathbhupendra
C
// A C program to check if there is a triplet with sum equal to 0 in
// a given BST
#include<stdio.h>
// A BST node has key, and left and right pointers
struct node
{
int key;
struct node *left;
struct node *right;
};
// A function to convert given BST to Doubly Linked List. left pointer is used
// as previous pointer and right pointer is used as next pointer. The function
// sets *head to point to first and *tail to point to last node of converted DLL
void convertBSTtoDLL(node* root, node** head, node** tail)
{
// Base case
if (root == NULL)
return;
// First convert the left subtree
if (root->left)
convertBSTtoDLL(root->left, head, tail);
// Then change left of current root as last node of left subtree
root->left = *tail;
// If tail is not NULL, then set right of tail as root, else current
// node is head
if (*tail)
(*tail)->right = root;
else
*head = root;
// Update tail
*tail = root;
// Finally, convert right subtree
if (root->right)
convertBSTtoDLL(root->right, head, tail);
}
// This function returns true if there is pair in DLL with sum equal
// to given sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
bool isPresentInDLL(node* head, node* tail, int sum)
{
while (head != tail)
{
int curr = head->key + tail->key;
if (curr == sum)
return true;
else if (curr > sum)
tail = tail->left;
else
head = head->right;
}
return false;
}
// The main function that returns true if there is a 0 sum triplet in
// BST otherwise returns false
bool isTripletPresent(node *root)
{
// Check if the given BST is empty
if (root == NULL)
return false;
// Convert given BST to doubly linked list. head and tail store the
// pointers to first and last nodes in DLLL
node* head = NULL;
node* tail = NULL;
convertBSTtoDLL(root, &head, &tail);
// Now iterate through every node and find if there is a pair with sum
// equal to -1 * head->key where head is current node
while ((head->right != tail) && (head->key < 0))
{
// If there is a pair with sum equal to -1*head->key, then return
// true else move forward
if (isPresentInDLL(head->right, tail, -1*head->key))
return true;
else
head = head->right;
}
// If we reach here, then there was no 0 sum triplet
return false;
}
// A utility function to create a new BST node with key as given num
node* newNode(int num)
{
node* temp = new node;
temp->key = num;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to insert a given key to BST
node* insert(node* root, int key)
{
if (root == NULL)
return newNode(key);
if (root->key > key)
root->left = insert(root->left, key);
else
root->right = insert(root->right, key);
return root;
}
// Driver program to test above functions
int main()
{
node* root = NULL;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root))
printf("Present");
else
printf("Not Present");
return 0;
}
Java
// A Java program to check if there
// is a triplet with sum equal to 0 in
// a given BST
import java.util.*;
class GFG
{
// A BST node has key, and left and right pointers
static class node
{
int key;
node left;
node right;
};
static node head;
static node tail;
// A function to convert given BST to Doubly
// Linked List. left pointer is used
// as previous pointer and right pointer
// is used as next pointer. The function
// sets *head to point to first and *tail
// to point to last node of converted DLL
static void convertBSTtoDLL(node root)
{
// Base case
if (root == null)
return;
// First convert the left subtree
if (root.left != null)
convertBSTtoDLL(root.left);
// Then change left of current root
// as last node of left subtree
root.left = tail;
// If tail is not null, then set right
// of tail as root, else current
// node is head
if (tail != null)
(tail).right = root;
else
head = root;
// Update tail
tail = root;
// Finally, convert right subtree
if (root.right != null)
convertBSTtoDLL(root.right);
}
// This function returns true if there
// is pair in DLL with sum equal to given
// sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
static boolean isPresentInDLL(node head, node tail, int sum)
{
while (head != tail)
{
int curr = head.key + tail.key;
if (curr == sum)
return true;
else if (curr > sum)
tail = tail.left;
else
head = head.right;
}
return false;
}
// The main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
static boolean isTripletPresent(node root)
{
// Check if the given BST is empty
if (root == null)
return false;
// Convert given BST to doubly linked list. head and tail store the
// pointers to first and last nodes in DLLL
head = null;
tail = null;
convertBSTtoDLL(root);
// Now iterate through every node and
// find if there is a pair with sum
// equal to -1 * head.key where head is current node
while ((head.right != tail) && (head.key < 0))
{
// If there is a pair with sum
// equal to -1*head.key, then return
// true else move forward
if (isPresentInDLL(head.right, tail, -1*head.key))
return true;
else
head = head.right;
}
// If we reach here, then
// there was no 0 sum triplet
return false;
}
// A utility function to create
// a new BST node with key as given num
static node newNode(int num)
{
node temp = new node();
temp.key = num;
temp.left = temp.right = null;
return temp;
}
// A utility function to insert a given key to BST
static node insert(node root, int key)
{
if (root == null)
return newNode(key);
if (root.key > key)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
// Driver code
public static void main(String[] args)
{
node root = null;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root))
System.out.print("Present");
else
System.out.print("Not Present");
}
}
// This code is contributed by aashish1995
Python3
# A Python program to check if there
# is a triplet with sum equal to 0 in
# a given BST
# A BST Node has key, and left and right pointers
class Node:
def __init__(self):
self.key = 0;
self.left = None;
self.right = None;
head = Node();
tail = Node();
# A function to convert given BST to Doubly
# Linked List. left pointer is used
# as previous pointer and right pointer
# is used as next pointer. The function
# sets *head to point to first and *tail
# to point to last Node of converted DLL
def convertBSTtoDLL(root):
# Base case
if (root == None):
return;
global tail;
global head;
# First convert the left subtree
if (root.left != None):
convertBSTtoDLL(root.left);
# Then change left of current root
# as last Node of left subtree
root.left = tail;
# If tail is not None, then set right
# of tail as root, else current
# Node is head
if (tail != None):
(tail).right = root;
else:
head = root;
# Update tail
tail = root;
# Finally, convert right subtree
if (root.right != None):
convertBSTtoDLL(root.right);
# This function returns True if there
# is pair in DLL with sum equal to given
# sum. The algorithm is similar to hasArrayTwoCandidates()
# in method 1 of http:#tinyurl.com/dy6palr
def isPresentInDLL(head, tail, s):
while (head != tail):
curr = head.key + tail.key;
if (curr == s):
return True;
elif(curr > s):
tail = tail.left;
else:
head = head.right;
return False;
# The main function that returns
# True if there is a 0 sum triplet in
# BST otherwise returns False
def isTripletPresent(root):
# Check if the given BST is empty
if (root == None):
return False;
# Convert given BST to doubly linked list. head and tail store the
# pointers to first and last Nodes in DLLL
global tail;
global head;
head = None;
tail = None;
convertBSTtoDLL(root);
# Now iterate through every Node and
# find if there is a pair with sum
# equal to -1 * head.key where head is current Node
while ((head.right != tail) and (head.key < 0)):
# If there is a pair with sum
# equal to -1*head.key, then return
# True else move forward
if (isPresentInDLL(head.right, tail, -1 * head.key)):
return True;
else:
head = head.right;
# If we reach here, then
# there was no 0 sum triplet
return False;
# A utility function to create
# a new BST Node with key as given num
def newNode(num):
temp = Node();
temp.key = num;
temp.left = temp.right = None;
return temp;
# A utility function to insert a given key to BST
def insert(root, key):
if (root == None):
return newNode(key);
if (root.key > key):
root.left = insert(root.left, key);
else:
root.right = insert(root.right, key);
return root;
# Driver code
if __name__ == '__main__':
root = None;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root)):
print("Present");
else:
print("Not Present");
# This code is contributed by Rajput-Ji
C#
// A C# program to check if there
// is a triplet with sum equal to 0 in
// a given BST
using System;
public class GFG
{
// A BST node has key, and left and right pointers
public class node
{
public int key;
public node left;
public node right;
};
static node head;
static node tail;
// A function to convert given BST to Doubly
// Linked List. left pointer is used
// as previous pointer and right pointer
// is used as next pointer. The function
// sets *head to point to first and *tail
// to point to last node of converted DLL
static void convertBSTtoDLL(node root)
{
// Base case
if (root == null)
return;
// First convert the left subtree
if (root.left != null)
convertBSTtoDLL(root.left);
// Then change left of current root
// as last node of left subtree
root.left = tail;
// If tail is not null, then set right
// of tail as root, else current
// node is head
if (tail != null)
(tail).right = root;
else
head = root;
// Update tail
tail = root;
// Finally, convert right subtree
if (root.right != null)
convertBSTtoDLL(root.right);
}
// This function returns true if there
// is pair in DLL with sum equal to given
// sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
static bool isPresentInDLL(node head, node tail, int sum)
{
while (head != tail)
{
int curr = head.key + tail.key;
if (curr == sum)
return true;
else if (curr > sum)
tail = tail.left;
else
head = head.right;
}
return false;
}
// The main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
static bool isTripletPresent(node root)
{
// Check if the given BST is empty
if (root == null)
return false;
// Convert given BST to doubly linked list. head and tail store the
// pointers to first and last nodes in DLLL
head = null;
tail = null;
convertBSTtoDLL(root);
// Now iterate through every node and
// find if there is a pair with sum
// equal to -1 * head.key where head is current node
while ((head.right != tail) && (head.key < 0))
{
// If there is a pair with sum
// equal to -1*head.key, then return
// true else move forward
if (isPresentInDLL(head.right, tail, -1*head.key))
return true;
else
head = head.right;
}
// If we reach here, then
// there was no 0 sum triplet
return false;
}
// A utility function to create
// a new BST node with key as given num
static node newNode(int num)
{
node temp = new node();
temp.key = num;
temp.left = temp.right = null;
return temp;
}
// A utility function to insert a given key to BST
static node insert(node root, int key)
{
if (root == null)
return newNode(key);
if (root.key > key)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
// Driver code
public static void Main(String[] args)
{
node root = null;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root))
Console.Write("Present");
else
Console.Write("Not Present");
}
}
// This code is contributed by aashish1995
Javascript
<script>
// A JavaScript program to check if there
// is a triplet with sum equal to 0 in
// a given BST
// A BST node has key,
// and left and right pointers
class node {
constructor(val) {
this.key = val;
this.left = null;
this.right = null;
}
}
var head;
var tail;
// A function to convert given BST to Doubly
// Linked List. left pointer is used
// as previous pointer and right pointer
// is used as next pointer. The function
// sets *head to point to first and *tail
// to point to last node of converted DLL
function convertBSTtoDLL( root) {
// Base case
if (root == null)
return;
// First convert the left subtree
if (root.left != null)
convertBSTtoDLL(root.left);
// Then change left of current root
// as last node of left subtree
root.left = tail;
// If tail is not null, then set right
// of tail as root, else current
// node is head
if (tail != null)
(tail).right = root;
else
head = root;
// Update tail
tail = root;
// Finally, convert right subtree
if (root.right != null)
convertBSTtoDLL(root.right);
}
// This function returns true if there
// is pair in DLL with sum equal to given
// sum. The algorithm is
// similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
function isPresentInDLL( head, tail , sum) {
while (head != tail) {
var curr = head.key + tail.key;
if (curr == sum)
return true;
else if (curr > sum)
tail = tail.left;
else
head = head.right;
}
return false;
}
// The main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
function isTripletPresent( root) {
// Check if the given BST is empty
if (root == null)
return false;
// Convert given BST to doubly
// linked list. head and tail store the
// pointers to first and last nodes in DLLL
head = null;
tail = null;
convertBSTtoDLL(root);
// Now iterate through every node and
// find if there is a pair with sum
// equal to -1 * head.key where
// head is current node
while ((head.right != tail) && (head.key < 0))
{
// If there is a pair with sum
// equal to -1*head.key, then return
// true else move forward
if (isPresentInDLL(head.right, tail,
-1 * head.key))
return true;
else
head = head.right;
}
// If we reach here, then
// there was no 0 sum triplet
return false;
}
// A utility function to create
// a new BST node with key as given num
function newNode(num) {
var temp = new node();
temp.key = num;
temp.left = temp.right = null;
return temp;
}
// A utility function to insert a given key to BST
function insert( root , key) {
if (root == null)
return newNode(key);
if (root.key > key)
root.left = insert(root.left, key);
else
root.right = insert(root.right, key);
return root;
}
// Driver code
var root = null;
root = insert(root, 6);
root = insert(root, -13);
root = insert(root, 14);
root = insert(root, -8);
root = insert(root, 15);
root = insert(root, 13);
root = insert(root, 7);
if (isTripletPresent(root))
document.write("Present");
else
document.write("Not Present");
// This code contributed by gauravrajput1
</script>
Producción
Present
Tenga en cuenta que la solución anterior modifica el BST dado.
Complejidad del tiempo: el tiempo necesario para convertir BST a DLL es O (n) y el tiempo necesario para encontrar el triplete en DLL es O (n ^ 2).
Espacio auxiliar: el espacio auxiliar solo se necesita para la pila de llamadas de funciones en la función recursiva convertBSTtoDLL(). Dado que el árbol dado está equilibrado (la altura es O (Inicio de sesión)), la cantidad de funciones en la pila de llamadas nunca será mayor que O (Inicio de sesión).
También podemos encontrar triplete en mismo tiempo y espacio extra sin modificar el árbol. Ver siguiente publicación. El código discutido allí también se puede usar para encontrar trillizos.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA