// 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;
//Shows a BST node which has key, and left and right pointers
class node{
public:
int key1;
node *left1;
node *right1;
};
// Shows 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 *head1 to point to first and *tail1
// to point to last node of converted DLL
void convertBSTtoDLL(node* root1, node** head1, node** tail1){
//Shows base case
if (root1 == NULL)
return;
// Now at first convert the left subtree
if (root1->left1)
convertBSTtoDLL(root1->left1, head1, tail1);
// After that change left of current root
// as last node of left subtree
root1->left1 = *tail1;
// Now if tail1 is not NULL, then set right
// of tail1 as root, else current
// node is head
if (*tail1)
(*tail1)->right1 = root1;
else
*head1 = root1;
// Used to update tail1
*tail1 = root1;
if (root1->right1)
convertBSTtoDLL(root1->right1, head1, tail1);
}
//Now this function returns true if there
// is pair in DLL with sum equal to given
// sum.
bool isPresentInDLL(node* head1, node* tail1, int sum1){
while (head1 != tail1){
int curr1 = head1->key1 + tail1->key1;
if (curr1 == sum1)
return true;
else if (curr1 > sum1)
tail1 = tail1->left1;
else
head1 = head1->right1;
}
return false;
}
// Now we explain the main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
bool isTripletPresent(node *root1){
// Verify if the given BST is empty
if (root1 == NULL)
return false;
// Now convert given BST to doubly linked list. head and tail store the
// pointers to first and last nodes in DLL
node* head1 = NULL;
node* tail1 = NULL;
convertBSTtoDLL(root1, &head1, &tail1);
// Now iterate through every node and
// determine if there is a pair with sum
// equal to -1 * head1->key1 where head1 is current node
while ((head1->right1 != tail1) && (head1->key1 < 0)){
// Now if there is a pair with sum
// equal to -1*head1->key1, then return
// true else move forward
if (isPresentInDLL(head1->right1, tail1, -1*head1->key1))
return true;
else
head1 = head1->right1;
}
// Now if we reach here, then
// there was no 0 sum triplet
return false;
}
// Shows a utility function to create
// a new BST node with key as given num
node* newNode(int num1){
node* temp1 = new node();
temp1->key1 = num1;
temp1->left1 = temp1->right1 = NULL;
return temp1;
}
//Shows a utility function to insert a given key to BST
node* insert(node* root1, int key1){
if (root1 == NULL)
return newNode(key1);
if (root1->key1 > key1)
root1->left1 = insert(root1->left1, key1);
else
root1->right1 = insert(root1->right1, key1);
return root1;
}
// Driver code
int main(){
node* root1 = NULL;
root1 = insert(root1, 7);
root1 = insert(root1, -15);
root1 = insert(root1, 15);
root1 = insert(root1, -7);
root1 = insert(root1, 14);
root1 = insert(root1, 16);
root1 = insert(root1, 8);
if (isTripletPresent(root1))
cout << "Present";
else
cout << "Not Present";
return 0;
}
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