Count BST nodes that lie in a given range

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

Examples:

Input:
        10
      /    \
    5       50
   /       /  \
 1       40   100
Range: [5, 45]

Output:  3
There are three nodes in range, 5, 10 and 40

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.



Below is the implementation of above idea.

C++

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// C++ program to count BST nodes withing 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;
}

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Java

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// 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

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Python3

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# Python3 program to count BST nodes 
# withing 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

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C#

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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

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Output:

Count of nodes between [5, 45] is 3

Time complexity of the above program is O(h + k) where h is height of BST and k is number of nodes in given range.

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.



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