BST in-order Traversal (Recursive Code and Complexity)

Pre-condition:

typedef struct intnode {
 int info;
 node *left, *right; 
}node;

Algorithm:
step 1: if pointer is null, return;
step 2: visit left sub-tree
step 3: visit root
step 4: visit right sub-tree
step 5: stop

Code:

void inorder(node *p) {
 // if p is null return
 if( p == null )
  return;
 // left sub-tree
 inorder(p->left);
 // root
 printf("%d", p->info);
 //right sub-tree
 inorder(p->right);
}

Complexity:
O(n), because each node has to be visited atleast once and increases linearly for increasing 'n'.