Question

Assume the following notation/operations on AVL trees. An empty AVL tree is denoted E. A non-empty AVL tree T has three attributes:

• The key T.key is the root node’s key.

• The left child T.left is T’s left subtree, which is an AVL tree (possibly E).

• The right child T.right is T’s right subtree, which is an AVL tree (possibly E).

Describe an alternative version of the RangeCount(T, lo, hi) function that runs in O(log n) time.

Answer 1

The idea is to traverse the given binary search tree starting from root. For every node being visited, check if this node lies in range, if yes, then add 1 to result and recur for both of its children. If current node is smaller than low value of range, then recur for right child, else recur for left child.

Since you have not mentioned the language of your preference, I am providing the code in C++.

CODE

// Returns count of nodes in BST in range [low, high]

int RangeCount(node *root, int low, int high)

{

    // Base case

    if (!root) return 0;

  

    // Special Optional case for improving efficiency

    if (root->data == high && root->data == low)

        return 1;

  

    // If current node is in range, then include it in count and

    // recur for left and right children of it

    if (root->data <= high && root->data >= low)

         return 1 + RangeCount(root->left, low, high) +

                    RangeCount(root->right, low, high);

  

    // If current node is smaller than low, then recur for right

    // child

    else if (root->data < low)

         return RangeCount(root->right, low, high);

  

    // Else recur for left child

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

}

Time complexity:

Time complexity of the above program is O(log n).