Find a pair with given sum in a Balanced BST

PROBLEM :

Given a Balanced Binary Search Tree and a target sum, write a function that returns true if there is a pair with sum equals to target sum, otherwise return false. Expected time complexity is O(n)
Any modification to Binary Search Tree is not allowed. Note that height of a Balanced BST is always O(Logn).

Example :
 for the above tree if user inputed target sum as 35 the output will be sum of 10,25 and 15,20
 
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SIMPLE c++ IMPLEMENTATION :
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The Brute Force Solution is to consider each pair in BST and check whether the sum equals to X. The time complexity of this solution will be O(n^2).

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 pair_sum_BST(tree *root,int ele)
{
int s,i ;
s=size_tree(root) ;
int array[s] ;
i=0 ;
inorder_array_fill(root,array,&i) ;

int a,b ;
a=0,b=s-1 ;
while(a<b)
{
if(array[a]+array[b]<ele)
a++ ;
else if(array[a]+array[b]>ele)
b-- ;
else
{
cout<<"\n sum pair found as "<<array[a]<<" and "<<array[b] ;
a++ ;
                        b-- ;
}
}
cout<<"\n No such sum pair found " ;
}

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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