BST中大于给定KEY的节点数

Question

I wrote the code bellow to count number of nodes in a BST which are bigger than a given KEY:

int Tree::findNumbersThatBiggerThanKey(int key){
    return PrivateFindNumbersThatBiggerThanKey(key, this->rootNode);
}

int Tree::PrivateFindNumbersThatBiggerThanKey(int key, Node* start) {
    if (!start)
        return 0;

    if (start->key > key)
        return 1 + this->PrivateFindNumbersThatBiggerThanKey(key, start->leftNode) +
        this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

    else if (start->key < key)
        return this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

    else /*start->key > key*/
        return this->PrivateFindNumbersThatBiggerThanKey(key, start->leftNode);
}

this is my main:

int main() {

    int TreeKeys[16] = { 50, 76, 21, 4, 32, 64, 15, 52, 14, 100, 83, 2, 3, 70, 87, 80 };
    Tree myBst;

    cout << "Printing the three Inorder before adding numbers: \n";
    myBst.printInorder();


    for (int i = 0; i < 16; i++) {
        myBst.AddLeaf(TreeKeys[i]);
    }

    cout << "Printing the three Inorder after adding numbers: \n";
    myBst.printInorder();

    int key = 2;
    cout << "\n\nNumber of nodes that are bigger than "<<key<<" : " <<
         myBst.findNumbersThatBiggerThanKey(key);
    return 0;
}

The Inroder print works just fine:

Printing the three Inorder after adding numbers:
2 3 4 14 15 21 32 50 52 64 70 76 80 83 87 100

but when I serach for number of nodes that are bigger than "2" I get 14 which is incorrect (instead of 15):

Number of nodes that are bigger than 2 : 14

When I search for number of nodes that are bigger than "1" I get 16 which is correct result (I get a correct result for any other numbers too):

Number of nodes that are bigger than 1 : 16

I'm a student and newbie with recursion, I'll be glad for explanations why that's happened and how can I fix it.

Answer 1

your else is not correct, it's condition is start->key == key which should be treated the same as start->key < key

so rewrite it to

else 
   return this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

or simply combine it with start->key < key

Answer 2

Since you don't provide an MCVE, I cannot work on your code. The general way of doing this, is like that:

int Tree::PrivateFindNumbersThatBiggerThanKey(int key, Node* node)
{
  if (node == null)
  {
    return 0;
  }
  int countLeft = PrivateFindNumbersThatBiggerThanKey(node->leftNode, key);
  int countRight = PrivateFindNumbersThatBiggerThanKey(node->rightNode, key);

  return (node->key > k ? 1 : 0) + countLeft + countRight;
}