Maximum number of leaf nodes that can be visited within the given budget

Last Updated : 18 Aug, 2021

Given a binary tree and an integer b representing budget. The task is to find the count of maximum number of leaf nodes that can be visited under the given budget if cost of visiting a leaf node is equal to level of that leaf node.
Note: Root of the tree is at level 1.
Examples:

Input: b = 8
           10
         /    \
        8       15
       /      /   \
      3      11     18 
              \
               13
Output: 2
For the above binary tree, leaf nodes are 3, 
13 and 18 at levels 3, 4 and 3 respectively.
Cost for visiting leaf node 3 is 3
Cost for visiting leaf node 13 is 4
Cost for visiting leaf node 18 is 3
Thus with given budget = 8, we can at maximum
visit two leaf nodes.

Input: b = 1
           8
         /   \
        7     10
       /      
      3      
             
Output: 0 
For the above binary tree, leaf nodes are
3 and 10 at levels 3 and 2 respectively.
Cost for visiting leaf node 3 is 3
Cost for visiting leaf node 10 is 2
In given budget = 1, we can't visit
any leaf node.

Approach:

Traverse the binary tree using level order traversal, and store the level of all the leaf nodes in a priority queue.
Remove an element from the priority queue one by one, check if this value is within the budget.
If Yes then subtract this value from the budget and update count = count + 1.
Else, print the count of maximum leaf nodes that can be visited within the given budget.

Below is the implementation of the above approach:

C++

// C++ program to calculate the maximum number of leaf
// nodes that can be visited within the given budget
#include <bits/stdc++.h>
using namespace std;

// struct that represents a node of the tree
struct Node {
    Node* left;
    Node* right;
    int data;
 
    Node(int key)
    {
        data = key;
        this->left = nullptr;
        this->right = nullptr;
    }
};

Node* newNode(int key)
{
    Node* temp = new Node(key);
    return temp;
}
 
// Priority queue to store the levels
// of all the leaf nodes
vector<int> pq;

// Level order traversal of the binary tree
void levelOrder(Node* root)
{
    vector<Node*> q;
    int len, level = 0;
    Node* temp;

    // If tree is empty
    if (root == nullptr)
        return;

    q.push_back(root);

    while (true) {

        len = q.size();
        if (len == 0)
            break;
        level++;
        while (len > 0) {

            temp = q[0];
            q.erase(q.begin());

            // If left child exists
            if (temp->left != nullptr)
                q.push_back(temp->left);

            // If right child exists
            if (temp->right != nullptr)
                q.push_back(temp->right);

            // If node is a leaf node
            if (temp->left == nullptr && temp->right == nullptr)
            {
                pq.push_back(level);
                sort(pq.begin(), pq.end());
                reverse(pq.begin(), pq.end());
            }
            len--;
        }
    }
}

// Function to calculate the maximum number of leaf nodes
// that can be visited within the given budget
int countLeafNodes(Node* root, int budget)
{
    levelOrder(root);
    int val;

    // Variable to store the count of
    // number of leaf nodes possible to visit
    // within the given budget
    int count = 0;

    while (pq.size() != 0) {

        // Removing element from
        // min priority queue one by one
        val = pq[0];
        pq.erase(pq.begin());

        // If current val is under budget, the
        // node can be visited
        // Update the budget afterwards
        if (val <= budget) {
            count++;
            budget -= val;
        }
        else
            break;
    }
    return count;
}

// Driver code
int main()
{
    Node* root = newNode(10);
    root->left = newNode(8);
    root->right = newNode(15);
    root->left->left = newNode(3);
    root->left->left->right = newNode(13);
    root->right->left = newNode(11);
    root->right->right = newNode(18);
 
    int budget = 8;
 
    cout << countLeafNodes(root, budget);

    return 0;
}

// This code is contributed by decode2207.

Java

// Java program to calculate the maximum number of leaf
// nodes that can be visited within the given budget
import java.io.*;
import java.util.*;
import java.lang.*;

// Class that represents a node of the tree
class Node {
    int data;
    Node left, right;

    // Constructor to create a new tree node
    Node(int key)
    {
        data = key;
        left = right = null;
    }
}

class GFG {

    // Priority queue to store the levels
    // of all the leaf nodes
    static PriorityQueue<Integer> pq;

    // Level order traversal of the binary tree
    static void levelOrder(Node root)
    {
        Queue<Node> q = new LinkedList<>();
        int len, level = 0;
        Node temp;

        // If tree is empty
        if (root == null)
            return;

        q.add(root);

        while (true) {

            len = q.size();
            if (len == 0)
                break;
            level++;
            while (len > 0) {

                temp = q.remove();

                // If left child exists
                if (temp.left != null)
                    q.add(temp.left);

                // If right child exists
                if (temp.right != null)
                    q.add(temp.right);

                // If node is a leaf node
                if (temp.left == null && temp.right == null)
                    pq.add(level);
                len--;
            }
        }
    }

    // Function to calculate the maximum number of leaf nodes
    // that can be visited within the given budget
    static int countLeafNodes(Node root, int budget)
    {
        pq = new PriorityQueue<>();
        levelOrder(root);
        int val;

        // Variable to store the count of
        // number of leaf nodes possible to visit
        // within the given budget
        int count = 0;

        while (pq.size() != 0) {

            // Removing element from
            // min priority queue one by one
            val = pq.poll();

            // If current val is under budget, the
            // node can be visited
            // Update the budget afterwards
            if (val <= budget) {
                count++;
                budget -= val;
            }
            else
                break;
        }
        return count;
    }

    // Driver code
    public static void main(String args[])
    {

        Node root = new Node(10);
        root.left = new Node(8);
        root.right = new Node(15);
        root.left.left = new Node(3);
        root.left.left.right = new Node(13);
        root.right.left = new Node(11);
        root.right.right = new Node(18);

        int budget = 8;

        System.out.println(countLeafNodes(root, budget));
    }
}

Python3

# Python3 program to calculate the maximum number of leaf
# nodes that can be visited within the given budget

# struct that represents a node of the tree
class Node:
    # Constructor to set the data of
    # the newly created tree node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

# Priority queue to store the levels
# of all the leaf nodes
pq = []

# Level order traversal of the binary tree
def levelOrder(root):

    q = []
    level = 0

    # If tree is empty
    if (root == None):
        return

    q.append(root)

    while (True) :

        Len = len(q)
        if (Len == 0):
            break
        level+=1
        while (Len > 0) :

            temp = q[0]
            del q[0]

            # If left child exists
            if (temp.left != None):
                q.append(temp.left)

            # If right child exists
            if (temp.right != None):
                q.append(temp.right)

            # If node is a leaf node
            if (temp.left == None and temp.right == None):
            
                pq.append(level)
                pq.sort()
                pq.reverse()
            
            Len-=1
    
    return pq

# Function to calculate the maximum number of leaf nodes
# that can be visited within the given budget
def countLeafNodes(root, budget):

    pq = levelOrder(root)

    # Variable to store the count of
    # number of leaf nodes possible to visit
    # within the given budget
    count = 0

    while (len(pq) != 0) :

        # Removing element from
        # min priority queue one by one
        val = pq[0]
        del pq[0]

        # If current val is under budget, the
        # node can be visited
        # Update the budget afterwards
        if (val <= budget) :
            count+=1
            budget -= val
        
        else:
            break
        
    return count

root = Node(10)
root.left = Node(8)
root.right = Node(15);
root.left.left = Node(3)
root.left.left.right = Node(13)
root.right.left = Node(11)
root.right.right = Node(18)

budget = 8

print(countLeafNodes(root, budget))

# This code is contributed by suresh07.

// C# program to calculate the maximum number of leaf
// nodes that can be visited within the given budget
using System;
using System.Collections.Generic;
class GFG {
    
    // Class that represents a node of the tree
    class Node
    {
        public Node left, right;
        public int data;
    };
    
    static Node newNode(int key)
    {
        Node temp = new Node();
        temp.data = key;
        temp.left = temp.right = null;
        return temp;
    }
    
    // Priority queue to store the levels
    // of all the leaf nodes
    static List<int> pq;
 
    // Level order traversal of the binary tree
    static void levelOrder(Node root)
    {
        List<Node> q = new List<Node>();
        int len, level = 0;
        Node temp;
 
        // If tree is empty
        if (root == null)
            return;
 
        q.Add(root);
 
        while (true) {
 
            len = q.Count;
            if (len == 0)
                break;
            level++;
            while (len > 0) {
 
                temp = q[0];
                q.RemoveAt(0);
 
                // If left child exists
                if (temp.left != null)
                    q.Add(temp.left);
 
                // If right child exists
                if (temp.right != null)
                    q.Add(temp.right);
 
                // If node is a leaf node
                if (temp.left == null && temp.right == null)
                {
                    pq.Add(level);
                    pq.Sort();
                    pq.Reverse();
                }
                len--;
            }
        }
    }
 
    // Function to calculate the maximum number of leaf nodes
    // that can be visited within the given budget
    static int countLeafNodes(Node root, int budget)
    {
        pq = new List<int>();
        levelOrder(root);
        int val;
 
        // Variable to store the count of
        // number of leaf nodes possible to visit
        // within the given budget
        int count = 0;
 
        while (pq.Count != 0) {
 
            // Removing element from
            // min priority queue one by one
            val = pq[0];
            pq.RemoveAt(0);
 
            // If current val is under budget, the
            // node can be visited
            // Update the budget afterwards
            if (val <= budget) {
                count++;
                budget -= val;
            }
            else
                break;
        }
        return count;
    }

  static void Main() {
    Node root = newNode(10);
    root.left = newNode(8);
    root.right = newNode(15);
    root.left.left = newNode(3);
    root.left.left.right = newNode(13);
    root.right.left = newNode(11);
    root.right.right = newNode(18);

    int budget = 8;

    Console.Write(countLeafNodes(root, budget));
  }
}

// This code is contributed by divyeshrabadiya07.

JavaScript

<script>

    // JavaScript program to calculate
    // the maximum number of leaf
    // nodes that can be visited within the given budget
    
    // Class that represents a node of the tree
    class Node {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
    
    // Priority queue to store the levels
    // of all the leaf nodes
    let pq = [];
  
    // Level order traversal of the binary tree
    function levelOrder(root)
    {
        let q = [];
        let len, level = 0;
        let temp;
  
        // If tree is empty
        if (root == null)
            return;
  
        q.push(root);
  
        while (true) {
  
            len = q.length;
            if (len == 0)
                break;
            level++;
            while (len > 0) {
  
                temp = q.shift();
  
                // If left child exists
                if (temp.left != null)
                    q.push(temp.left);
  
                // If right child exists
                if (temp.right != null)
                    q.push(temp.right);
  
                // If node is a leaf node
                if (temp.left == null && temp.right == null)
                {
                    pq.push(level);
                    pq.sort(function(a, b){return a - b});
                    pq.reverse();
                }
                len--;
            }
        }
    }
  
    // Function to calculate the maximum number of leaf nodes
    // that can be visited within the given budget
    function countLeafNodes(root, budget)
    {
        pq = [];
        levelOrder(root);
        let val;
  
        // Variable to store the count of
        // number of leaf nodes possible to visit
        // within the given budget
        let count = 0;
  
        while (pq.length != 0) {
  
            // Removing element from
            // min priority queue one by one
            val = pq[0];
            pq.shift();
  
            // If current val is under budget, the
            // node can be visited
            // Update the budget afterwards
            if (val <= budget) {
                count++;
                budget -= val;
            }
            else
                break;
        }
        return count;
    }
    
    let root = new Node(10);
    root.left = new Node(8);
    root.right = new Node(15);
    root.left.left = new Node(3);
    root.left.left.right = new Node(13);
    root.right.left = new Node(11);
    root.right.right = new Node(18);

    let budget = 8;

    document.write(countLeafNodes(root, budget));

</script>

Output:

Level with maximum number of nodes using DFS in a N-ary tree

rachana soma

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Maximum number of leaf nodes that can be visited within the given budget

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