Find if there is a triplet in a Balanced BST that adds to zero

PROBLEM :

Given a Balanced Binary Search Tree (BST), write a function isTripletPresent() that returns true if there is a triplet in given BST with sum equals to 0, otherwise returns false. Expected time complexity is O(n^2) . You can modify given Binary Search Tree. Note that height of a Balanced BST is always O(Logn)

For example, isTripletPresent() should return true for following BST because there is a triplet with sum 0, the triplet is {-13, 6, 7}.

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SIMPLE c++ IMPLEMENTATION :
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The Brute Force Solution is to consider each triplet in BST and check whether the sum adds upto zero. The time complexity of this solution will be O(n^3).

BETTER SOLUTION :

/*
typedef struct BST
{
int info ;
struct BST *left ;
struct BST *right ;
}tree ;                            */

int size_tree(tree*) ;
void inorder_array_fill(tree* ,int [],int *) ;

void triplet_tree(tree *root)
{
int h,i ;
h=size_tree(root) ;
i=0 ;
int array[h] ;
inorder_array_fill(root,array,&i) ;

int a,b,c ;

for(i=0;i<h;i++)
{
a=array[i] ;
b=i+1 ;
c=h-1 ;
while(b<c)
{
if(a+array[b]+array[c]==0)
{
cout<<" Triplate found !!! elements are "<<a<<" "<<array[b]<<" "<<array[c] ;
break ;
}
else if(a+array[b]+array[c]<0)
b++ ;
else
c-- ;

}
}

cout<<" \n No Triplate preasent " ;
}


int size_tree(tree* root)
{
if(root==NULL)
return 0 ;

return ((size_tree(root->left))+(size_tree(root->right))+1) ;
}

void inorder_array_fill(tree* root,int a[],int *s)
{
    if(root==NULL)
        return  ;
     
    inorder_array_fill(root->left,a,&(*s)) ;
    a[(*s)++]=root->info ;
    inorder_array_fill(root->right,a,&(*s)) ;
 
}

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