Count BST subtrees that lie in given range

Given a Binary Search Tree (BST) of integer values and a range [low, high], return the count of nodes where all the nodes under that node (or subtree rooted with that node) lie in the given range. 

Examples:

Input:
        10
      /    \
    5       50
   /       /  \
 1       40   100
Range: [5, 45]
Output:  1 
There is only 1 node whose subtree is in the given range.
The node is 40 


Input:
        10
      /    \
    5       50
   /       /  \
 1       40   100
Range: [1, 45]
Output:  3 
There are three nodes whose subtree is in the given range.
The nodes are 1, 5 and 40 

We strongly recommend you minimize your browser and try this yourself first.
The idea is to traverse the given Binary Search Tree (BST) in a bottom-up manner. For every node, recur for its subtrees, if subtrees are in range and the nodes are also in range, then increment the count and return true (to tell the parent about its status). The count is passed as a pointer so that it can be incremented across all function calls.

Below is the implementation of the above idea. 

C++14

// C++ program to count subtrees that lie in a given range
#include <bits/stdc++.h>
using namespace std;

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

// A utility function to check if data of root is
// in range from low to high
bool inRange(node* root, int low, int high)
{
    return root->data >= low && root->data <= high;
}

// A recursive function to get count of nodes whose subtree
// is in range from low to high.  This function returns true
// if nodes in subtree rooted under 'root' are in range.
bool getCountUtil(node* root, int low, int high, int* count)
{
    // Base case
    if (root == NULL)
        return true;

    // Recur for left and right subtrees
    bool l = getCountUtil(root->left, low, high, count);
    bool r = getCountUtil(root->right, low, high, count);

    // If both left and right subtrees are in range and current node
    // is also in range, then increment count and return true
    if (l && r && inRange(root, low, high)) {
        ++*count;
        return true;
    }

    return false;
}

// A wrapper over getCountUtil(). This function initializes count as 0
// and calls getCountUtil()
int getCount(node* root, int low, int high)
{
    int count = 0;
    getCountUtil(root, low, high, &count);
    return count;
}

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

// 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 subtrees in [" << l << ", "
         << h << "] is " << getCount(root, l, h);
    return 0;
}

Java

// Java program to count subtrees
// that lie in a given range
class GfG {

    // A BST node
    static class node {
        int data;
        node left, right;
    };

    // int class
    static class INT {
        int a;
    }

    // A utility function to check if data of root is
    // in range from low to high
    static boolean inRange(node root, int low, int high)
    {
        return root.data >= low && root.data <= high;
    }

    // A recursive function to get count
    // of nodes whose subtree is in range
    // from low to high. This function returns
    // true if nodes in subtree rooted under
    // 'root' are in range.
    static boolean getCountUtil(node root, int low,
                                int high, INT count)
    {
        // Base case
        if (root == null)
            return true;

        // Recur for left and right subtrees
        boolean l = getCountUtil(root.left,
                                 low, high, count);
        boolean r = getCountUtil(root.right,
                                 low, high, count);

        // If both left and right subtrees are
        // in range and current node is also in
        // range, then increment count and return true
        if (l && r && inRange(root, low, high)) {
            ++count.a;
            return true;
        }

        return false;
    }

    // A wrapper over getCountUtil().
    // This function initializes count as 0
    // and calls getCountUtil()
    static INT getCount(node root, int low, int high)
    {
        INT count = new INT();
        count.a = 0;
        getCountUtil(root, low, high, count);
        return count;
    }

    // Utility function to create new node
    static node newNode(int data)
    {
        node temp = new node();
        temp.data = data;
        temp.left = temp.right = null;
        return (temp);
    }

    // Driver code
    public static void main(String args[])
    {
        // 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 construct BST shown in above example
        10
        / \
    5 50
    / / \
    1 40 100 */
        int l = 5;
        int h = 45;
        System.out.println("Count of subtrees in [" + l + ", "
                           + h + "] is " + getCount(root, l, h).a);
    }
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to count subtrees that
# lie in 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

# A utility function to check if data of
# root is in range from low to high
def inRange(root, low, high):
    return root.data >= low and root.data <= high

# A recursive function to get count of nodes
# whose subtree is in range from low to high.
# This function returns true if nodes in subtree
# rooted under 'root' are in range.
def getCountUtil(root, low, high, count):

    # Base case
    if root == None:
        return True

    # Recur for left and right subtrees
    l = getCountUtil(root.left, low, high, count)
    r = getCountUtil(root.right, low, high, count)

    # If both left and right subtrees are in range
    # and current node is also in range, then
    # increment count and return true
    if l and r and inRange(root, low, high):
        count[0] += 1
        return True

    return False

# A wrapper over getCountUtil(). This function
# initializes count as 0 and calls getCountUtil()
def getCount(root, low, high):
    count = [0]
    getCountUtil(root, low, high, count)
    return count

# 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 subtrees in [", l, ", ", h, "] is ",
                                   getCount(root, l, h))

# This code is contributed by PranchalK

C#

// C# program to count subtrees
// that lie in a given range
using System;

class GfG {

    // A BST node
    public class node {
        public int data;
        public node left, right;
    };

    // int class
    public class INT {
        public int a;
    }

    // A utility function to check if data of root is
    // in range from low to high
    static bool inRange(node root, int low, int high)
    {
        return root.data >= low && root.data <= high;
    }

    // A recursive function to get count
    // of nodes whose subtree is in range
    // from low to high. This function returns
    // true if nodes in subtree rooted under
    // 'root' are in range.
    static bool getCountUtil(node root, int low,
                             int high, INT count)
    {
        // Base case
        if (root == null)
            return true;

        // Recur for left and right subtrees
        bool l = getCountUtil(root.left,
                              low, high, count);
        bool r = getCountUtil(root.right,
                              low, high, count);

        // If both left and right subtrees are
        // in range and current node is also in
        // range, then increment count and return true
        if (l && r && inRange(root, low, high)) {
            ++count.a;
            return true;
        }

        return false;
    }

    // A wrapper over getCountUtil().
    // This function initializes count as 0
    // and calls getCountUtil()
    static INT getCount(node root, int low, int high)
    {
        INT count = new INT();
        count.a = 0;
        getCountUtil(root, low, high, count);
        return count;
    }

    // Utility function to create new node
    static node newNode(int data)
    {
        node temp = new node();
        temp.data = data;
        temp.left = temp.right = null;
        return (temp);
    }

    // Driver code
    public static void Main(String[] args)
    {
        // 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 construct BST shown in above example
        10
        / \
    5 50
    / / \
    1 40 100 */
        int l = 5;
        int h = 45;
        Console.WriteLine("Count of subtrees in [" + l + ", "
                          + h + "] is " + getCount(root, l, h).a);
    }
}

// This code contributed by Rajput-Ji

Javascript

<script>

// JavaScript program to count subtrees
// that lie in a given range


    // A BST node

     class node {
            constructor(val) {
                this.data = val;
                this.left = null;
                this.right = null;
            }
        }


    // var class
     class INT {
         constructor(){
         this.a = 0;
         }
    }

    // A utility function to check if data of root is
    // in range from low to high
    function inRange( root , low , high)
    {
        return root.data >= low && root.data <= high;
    }

    // A recursive function to get count
    // of nodes whose subtree is in range
    // from low to high. This function returns
    // true if nodes in subtree rooted under
    // 'root' are in range.
    function getCountUtil( root , low,
                                 high,  count)
    {
        // Base case
        if (root == null)
            return true;

        // Recur for left and right subtrees
        var l = getCountUtil(root.left,
                                 low, high, count);
        var r = getCountUtil(root.right,
                                 low, high, count);

        // If both left and right subtrees are
        // in range and current node is also in
        // range, then increment count and return true
        if (l && r && inRange(root, low, high)) {
            ++count.a;
            return true;
        }

        return false;
    }

    // A wrapper over getCountUtil().
    // This function initializes count as 0
    // and calls getCountUtil()
     function getCount( root , low , high)
    {
        var count = new INT();
        count.a = 0;
        getCountUtil(root, low, high, count);
        return count;
    }

    // Utility function to create new node
     function newNode(data)
    {
        var temp = new node();
        temp.data = data;
        temp.left = temp.right = null;
        return (temp);
    }

    // Driver code

        // Let us construct  the BST shown in the above figure
        var 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 construct BST shown in above example
        10
        / \
    5 50
    / / \
    1 40 100 */
        var l = 5;
        var h = 45;
        document.write("Count of subtrees in [" + l + ", "
                    + h + "] is " + getCount(root, l, h).a);

// This code is contributed by todaysgaurav

</script>

Output

Count of subtrees in [5, 45] is 1

Time Complexity: O(n), As we are visiting every node just once. Here n is the number of nodes in the tree.
Auxiliary Space: O(h), where h is the height of the tree this is due to the recursive call stack.

if we exclude the recursion call stack space, otherwise the Auxiliary Space is O(h) where h is the height of the tree.