Given a Balanced Binary Search Tree (BST), the task is to write a function isTripletPresent() which returns true if there is a triplet in the given BST with a sum equal to 0, otherwise returns false.
The expected time complexity should be O(n^2) and only O(Logn) extra space can be used. You can modify the given Binary Search Tree. Note that the height of a Balanced BST is always O(Log n).
For example, isTripletPresent() should return true for following BST because there is a triplet with sum 0, the triplet is {-13, 6, 7}.
The Brute Force Solution is to consider each triplet in BST and check whether the sum adds up to zero. The time complexity of this solution will be O(n^3).
A Better Solution is to create an auxiliary array and store the Inorder traversal of BST in the array. The array will be sorted as Inorder traversal of BST always produces sorted data. Once we have the Inorder traversal, we can use method 2 of this post to find the triplet with a sum equals to 0. This solution works in O(n^2) time but requires O(n) auxiliary space.
Following is the solution that works in O(n^2) time and uses O(Logn) extra space:
- Convert given BST to Doubly Linked List (DLL)
- Now iterate through every node of DLL and if the key of node is negative, then find a pair in DLL with sum equal to key of current node multiplied by -1. To find the pair, we can use the approach used in hasArrayTwoCandidates() in method 1 of this post.
Implementation:
// 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 |
// 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;
} |
// 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 |
# 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 |
// 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 |
<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> |
Present
Note that the above solution modifies given BST.
Time Complexity: Time taken to convert BST to DLL is O(n) and time taken to find triplet in DLL is O(n^2).
Auxiliary Space: The auxiliary space is needed only for function call stack in recursive function convertBSTtoDLL(). Since given tree is balanced (height is O(Logn)), the number of functions in call stack will never be more than O(Logn).
We can also find triplet in same time and extra space without modifying the tree. See next post. The code discussed there can be used to find triplet also.