Count BST nodes that lie in a given range

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

Examples:

Input: Range: [5, 45]

         10
       /     \
    5       50
   /        /    \
 1      40   100
Output:  3
Explanation: There are three nodes in range, 5, 10 and 40

Input: Range: [10, 100]

          10
        /    \
     5      50
   /       /    \
 1     40   100
Output:  4

Approach:

Traverse the given binary search tree starting from root. For every node 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.

Follow the below steps to Implement the idea:

• Traverse the tree starting from the root.
  • If root->data is equal to high and root->data is equal to low return 1.
  • If root->data <= high and root->data >= low then return 1 + count on left of root + count on right of root.
  • Else if root->data < low return count on right of root.
  • Else if root->data > high return count on right of root.

Below is the implementation of the above approach.  

C++

// C++ program to count BST nodes within a given range
#include<bits/stdc++.h>
using namespace std;

// A BST node
struct node
{
    int data;
    struct node* left, *right;
};

// Utility function to create new node
node *newNode(int data)
{
    node *temp = new node;
    temp->data  = data;
    temp->left  = temp->right = NULL;
    return (temp);
}

// Returns count of nodes in BST in range [low, high]
int getCount(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 + getCount(root->left, low, high) +
                    getCount(root->right, low, high);

    // If current node is smaller than low, then recur for right
    // child
    else if (root->data < low)
         return getCount(root->right, low, high);

    // Else recur for left child
    else return getCount(root->left, low, high);
}

// Driver program
int main()
{
    // Let us construct the BST shown in the above figure
    node *root        = newNode(10);
    root->left        = newNode(5);
    root->right       = newNode(50);
    root->left->left  = newNode(1);
    root->right->left = newNode(40);
    root->right->right = newNode(100);
    /* Let us constructed BST shown in above example
          10
        /    \
      5       50
     /       /  \
    1       40   100   */
    int l = 5;
    int h = 45;
    cout << "Count of nodes between [" << l << ", " << h
         << "] is " << getCount(root, l, h);
    return 0;
}

Java

// Java code to count BST nodes that
// lie in a given range
class BinarySearchTree {

    /* Class containing left and right child
    of current node and key value*/
    static class Node {
        int data;
        Node left, right;

        public Node(int item) {
            data = item;
            left = right = null;
        }
    }

    // Root of BST
    Node root;

    // Constructor
    BinarySearchTree() {
        root = null;
    }

    // Returns count of nodes in BST in
    // range [low, high]
    int getCount(Node node, int low, int high)
    {
        // Base Case
        if(node == null)
            return 0;

        // If current node is in range, then
        // include it in count and recur for
        // left and right children of it
        if(node.data >= low && node.data <= high)
            return 1 + this.getCount(node.left, low, high)+
                this.getCount(node.right, low, high);

        // If current node is smaller than low,
        // then recur for right child
        else if(node.data < low)
            return this.getCount(node.right, low, high);

        // Else recur for left child
        else
            return this.getCount(node.left, low, high);
    }

    // Driver function
    public static void main(String[] args) {
        BinarySearchTree tree = new BinarySearchTree();

        tree.root = new Node(10);
        tree.root.left     = new Node(5);
        tree.root.right     = new Node(50);
        tree.root.left.left = new Node(1);
        tree.root.right.left = new Node(40);
        tree.root.right.right = new Node(100);
        /* Let us constructed BST shown in above example
          10
        /    \
      5       50
     /       /  \
    1       40   100   */
    int l=5;
    int h=45;
    System.out.println("Count of nodes between [" + l + ", "
                      + h+ "] is " + tree.getCount(tree.root,
                                                  l, h));
    }
}
// This code is contributed by Kamal Rawal

Python3

# Python3 program to count BST nodes
# within a given range

# Utility function to create new node
class newNode:

    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Returns count of nodes in BST in
# range [low, high]
def getCount(root, low, high):

    # Base case
    if root == None:
        return 0

    # Special Optional case for improving
    # efficiency
    if root.data == high and 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 and root.data >= low:
        return (1 + getCount(root.left, low, high) +
                    getCount(root.right, low, high))

    # If current node is smaller than low,
    # then recur for right child
    elif root.data < low:
        return getCount(root.right, low, high)

    # Else recur for left child
    else:
        return getCount(root.left, low, high)

# Driver Code
if __name__ == '__main__':

    # Let us construct the BST shown in
    # the above figure
    root = newNode(10)
    root.left = newNode(5)
    root.right = newNode(50)
    root.left.left = newNode(1)
    root.right.left = newNode(40)
    root.right.right = newNode(100)

    # Let us constructed BST shown in above example
    #     10
    #     / \
    # 5     50
    # /     / \
    # 1     40 100
    l = 5
    h = 45
    print("Count of nodes between [", l, ", ", h,"] is ",
                                    getCount(root, l, h))

# This code is contributed by PranchalK

C#

using System;

// C# code to count BST nodes that
// lie in a given range
public class BinarySearchTree
{

    /* Class containing left and right child
    of current node and key value*/
    public class Node
    {
        public int data;
        public Node left, right;

        public Node(int item)
        {
            data = item;
            left = right = null;
        }
    }

    // Root of BST
    public Node root;

    // Constructor
    public BinarySearchTree()
    {
        root = null;
    }

    // Returns count of nodes in BST in
    // range [low, high]
    public virtual int getCount(Node node, int low, int high)
    {
        // Base Case
        if (node == null)
        {
            return 0;
        }

        // If current node is in range, then
        // include it in count and recur for
        // left and right children of it
        if (node.data >= low && node.data <= high)
        {
            return 1 + this.getCount(node.left, low, high) + this.getCount(node.right, low, high);
        }

        // If current node is smaller than low,
        // then recur for right child
        else if (node.data < low)
        {
            return this.getCount(node.right, low, high);
        }

        // Else recur for left child
        else
        {
            return this.getCount(node.left, low, high);
        }
    }

    // Driver function
    public static void Main(string[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();

        tree.root = new Node(10);
        tree.root.left = new Node(5);
        tree.root.right = new Node(50);
        tree.root.left.left = new Node(1);
        tree.root.right.left = new Node(40);
        tree.root.right.right = new Node(100);
        /* Let us constructed BST shown in above example
          10
        /    \
      5       50
     /       /  \
    1       40   100   */
    int l = 5;
    int h = 45;
    Console.WriteLine("Count of nodes between [" + l + ", " + h + "] is " + tree.getCount(tree.root, l, h));
    }
}

  // This code is contributed by Shrikant13

Javascript

<script>
// javascript code to count BST nodes that
// lie in a given range


    /*
     * Class containing left and right child of current node and key value
     */
     class Node {
        constructor(item) {
            this.data = item;
            this.left =this.right = null;
        }
    }

    // Root of BST
    var root = null;



    // Returns count of nodes in BST in
    // range [low, high]
    function getCount( node , low , high) {
        // Base Case
        if (node == null)
            return 0;

        // If current node is in range, then
        // include it in count and recur for
        // left and right children of it
        if (node.data >= low && node.data <= high)
            return 1 + this.getCount(node.left, low, high) + this.getCount(node.right, low, high);

        // If current node is smaller than low,
        // then recur for right child
        else if (node.data < low)
            return this.getCount(node.right, low, high);

        // Else recur for left child
        else
            return this.getCount(node.left, low, high);
    }

    // Driver function



        root = new Node(10);
        root.left = new Node(5);
        root.right = new Node(50);
        root.left.left = new Node(1);
        root.right.left = new Node(40);
        root.right.right = new Node(100);
        /*
         * Let us constructed BST shown in above example 10 / \ 5 50 / / \ 1 40 100
         */
        var l = 5;
        var h = 45;
        document.write("Count of nodes between [" + l + ", " + h + "] is " + getCount(root, l, h));

// This code contributed by aashish1995
</script>

Output

Count of nodes between [5, 45] is 3

Time complexity: O(H + k) where h is the height of BST and k is the number of nodes in the given range.
Auxiliary Space: O(n) 

This article is contributed by Gaurav Ahirwar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.