inputs: Three numbers in a BST that adds upto zero | A C 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 ; 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 ; Base case ; First convert the left subtree ; Then change left of current root as last node of left subtree ; If tail is not NULL , then set right of tail as root , else current node is head ; Update tail ; Finally , convert right subtree ; This function returns true if there is pair in DLL with sum equal to given sum . The algorithm is similar to hasArrayTwoCandidates ( ) tinyurl . com / dy6palr in method 1 of http : ; The main function that returns true if there is a 0 sum triplet in BST otherwise returns false ; Check if the given BST is empty ; Convert given BST to doubly linked list . head and tail store the pointers to first and last nodes in DLLL ; Now iterate through every node and find if there is a pair with sum equal to - 1 * heaf -> key where head is current node ; If there is a pair with sum equal to - 1 * head -> key , then return true else move forward ; If we reach here , then there was no 0 sum triplet ; A utility function to create a new BST node with key as given num ; A utility function to insert a given key to BST ; Driver program to test above functions
How can the above be solved in C?
targets: #include <stdio.h>
struct node { int key ; struct node * left ; struct node * right ; } ; void convertBSTtoDLL ( node * root , node * * head , node * * tail ) { if ( root == NULL ) return ; if ( root -> left ) convertBSTtoDLL ( root -> left , head , tail ) ; root -> left = * tail ; if ( * tail ) ( * tail ) -> right = root ; else * head = root ; * tail = root ; if ( root -> right ) convertBSTtoDLL ( root -> right , head , tail ) ; } 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 ; } bool isTripletPresent ( node * root ) { if ( root == NULL ) return false ; node * head = NULL ; node * tail = NULL ; convertBSTtoDLL ( root , & head , & tail ) ; while ( ( head -> right != tail ) && ( head -> key < 0 ) ) { if ( isPresentInDLL ( head -> right , tail , -1 * head -> key ) ) return true ; else head = head -> right ; } return false ; } node * newNode ( int num ) { node * temp = new node ; temp -> key = num ; temp -> left = temp -> right = NULL ; return temp ; } 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 ; } 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 ; }