Count BST nodes that lie in a given range.cpp

/*
Count BST nodes that lie in a given range
==========================================

Given a Binary Search Tree (BST) and a range l-h(inclusive), count the number of nodes in the BST that lie in the given range.

The values smaller than root go to the left side
The values greater and equal to the root go to the right side

Example 1:
Input:
      10
     /  \
    5    50
   /    /  \
  1    40  100
l = 5, h = 45
Output: 3
Explanation: 5 10 40 are the node in the
range

Example 2:
Input:
     5
    /  \
   4    6
  /      \
 3        7
l = 2, h = 8
Output: 5
Explanation: All the nodes are in the
given range.

Your Task:
This is a function problem. You don't have to take input. You are required to complete the function getCountOfNode() that takes root, l ,h as parameters and returns the count.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(Height of the BST).

Constraints:
1 <= Number of nodes <= 100
1 <= l < h < 103
*/

int getCount(Node *root, int low, int high)
{
  if (!root)
    return 0;

  if (root->data == high && root->data == low)
    return 1;

  if (root->data <= high && root->data >= low)
    return 1 + getCount(root->left, low, high) +
           getCount(root->right, low, high);

  else if (root->data < low)
    return getCount(root->right, low, high);

  else
    return getCount(root->left, low, high);
}