Find if there is a path between two vertices in a directed graph

Level with maximum number of nodes

Last Updated : 25 Apr, 2025

Try it on GfG Practice

Given a binary tree, the task is to find the level in a binary tree that has the maximum number of nodes.

Note: The root is at level 0.

Examples:

Input: Binary Tree

Output : 2
Explanation:

Input: Binary tree

Output:1
Explanation

Using Breadth First Search - O(n) time and O(n) space

The idea is to traverse the binary tree level by level, keeping track of the number of nodes at each level. As we process each level, we compare its node count with the maximum count found so far, updating our result level if the current level has more nodes. The queue-based BFS naturally groups nodes by their level, making it straightforward to count nodes per level.

Step by step approach:

Start with root node in the queue and initialize tracking variables for level, max nodes, and result level.
For each level, record its size (number of nodes) from the queue before processing.
If current level's size exceeds previous maximum, update the maximum count and result level.
Process all nodes at current level, adding their children to queue for next level processing.
Return the level that had the maximum number of nodes after all levels are processed.

C++

// C++ code to find Level with maximum number of nodes using BFS
#include <iostream>
#include <queue>
using namespace std;

class Node {
public:
    int data;
    Node *left, *right;
    Node(int x) {
        data = x;
        left = nullptr;
        right = nullptr;
    }
};

int maxNodeLevel(Node *root) {
    if (root == nullptr)
        return -1;
    
    queue<Node*> q;
    q.push(root);
    
    int level = 0;
    int maxNodes = 0;
    int maxLevel = 0;
    
    while (!q.empty()) {
        
        // Number of nodes at current level
        int size = q.size();  
        
        if (size > maxNodes) {
            maxNodes = size;
            maxLevel = level;
        }
        
        // Process all nodes at current level
        for (int i = 0; i < size; i++) {
            Node* current = q.front();
            q.pop();
            
            if (current->left)
                q.push(current->left);
            if (current->right)
                q.push(current->right);
        }
        
        level++;
    }
    
    return maxLevel;
}

int main() {
    
    //          2
    //        /   \
    //       1     3
    //      / \     \
    //     4   6     8
    //    /
    //   5
    Node* root = new Node(2);
    root->left = new Node(1);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(6);
    root->right->right = new Node(8);
    root->left->left->left = new Node(5);
    
    cout <<  maxNodeLevel(root) << endl;
    
    return 0;
}

Java

// Java code to find Level with maximum number of nodes using BFS
import java.util.Queue;
import java.util.LinkedList;

class Node {
    int data;
    Node left, right;
    Node(int x) {
        data = x;
        left = null;
        right = null;
    }
}

class GfG {

    // A function to find level with maximum number of nodes
    static int maxNodeLevel(Node root) {
        if (root == null)
            return -1;

        Queue<Node> q = new LinkedList<>();
        q.add(root);

        int level = 0;
        int maxNodes = 0;
        int maxLevel = 0;

        while (!q.isEmpty()) {

            // Number of nodes at current level
            int size = q.size();  

            if (size > maxNodes) {
                maxNodes = size;
                maxLevel = level;
            }

            // Process all nodes at current level
            for (int i = 0; i < size; i++) {
                Node current = q.poll();

                if (current.left != null)
                    q.add(current.left);
                if (current.right != null)
                    q.add(current.right);
            }

            level++;
        }

        return maxLevel;
    }

    public static void main(String[] args) {

        //          2
        //        /   \
        //       1     3
        //      / \     \
        //     4   6     8
        //    /
        //   5
        Node root = new Node(2);
        root.left = new Node(1);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(6);
        root.right.right = new Node(8);
        root.left.left.left = new Node(5);

        System.out.println(maxNodeLevel(root));
    }
}

Python

# Python code to find Level with maximum number of nodes using BFS
from collections import deque

class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# A function to find level with maximum number of nodes
def maxNodeLevel(root):
    if root is None:
        return -1

    q = deque()
    q.append(root)

    level = 0
    maxNodes = 0
    maxLevel = 0

    while q:

        # Number of nodes at current level
        size = len(q)

        if size > maxNodes:
            maxNodes = size
            maxLevel = level

        # Process all nodes at current level
        for _ in range(size):
            current = q.popleft()

            if current.left:
                q.append(current.left)
            if current.right:
                q.append(current.right)

        level += 1

    return maxLevel

if __name__ == "__main__":

    #          2
    #        /   \
    #       1     3
    #      / \     \
    #     4   6     8
    #    /
    #   5
    root = Node(2)
    root.left = Node(1)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(6)
    root.right.right = Node(8)
    root.left.left.left = Node(5)

    print(maxNodeLevel(root))

// C# code to find Level with maximum number of nodes using BFS
using System;
using System.Collections.Generic;

class Node {
    public int data;
    public Node left, right;
    public Node(int x) {
        data = x;
        left = null;
        right = null;
    }
}

class GfG {

    // A function to find level with maximum number of nodes
    static int maxNodeLevel(Node root) {
        if (root == null)
            return -1;

        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);

        int level = 0;
        int maxNodes = 0;
        int maxLevel = 0;

        while (q.Count > 0) {

            // Number of nodes at current level
            int size = q.Count;

            if (size > maxNodes) {
                maxNodes = size;
                maxLevel = level;
            }

            // Process all nodes at current level
            for (int i = 0; i < size; i++) {
                Node current = q.Dequeue();

                if (current.left != null)
                    q.Enqueue(current.left);
                if (current.right != null)
                    q.Enqueue(current.right);
            }

            level++;
        }

        return maxLevel;
    }

    static void Main(string[] args) {

        //          2
        //        /   \
        //       1     3
        //      / \     \
        //     4   6     8
        //    /
        //   5
        Node root = new Node(2);
        root.left = new Node(1);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(6);
        root.right.right = new Node(8);
        root.left.left.left = new Node(5);

        Console.WriteLine(maxNodeLevel(root));
    }
}

JavaScript

// JavaScript code to find Level with maximum number of nodes using BFS

class Node {
    constructor(x) {
        this.data = x;
        this.left = null;
        this.right = null;
    }
}

// A function to find level with maximum number of nodes
function maxNodeLevel(root) {
    if (root === null)
        return -1;

    let q = [];
    q.push(root);

    let level = 0;
    let maxNodes = 0;
    let maxLevel = 0;

    while (q.length > 0) {

        // Number of nodes at current level
        let size = q.length;

        if (size > maxNodes) {
            maxNodes = size;
            maxLevel = level;
        }

        // Process all nodes at current level
        for (let i = 0; i < size; i++) {
            let current = q.shift();

            if (current.left !== null)
                q.push(current.left);
            if (current.right !== null)
                q.push(current.right);
        }

        level++;
    }

    return maxLevel;
}

//          2
//        /   \
//       1     3
//      / \     \
//     4   6     8
//    /
//   5
let root = new Node(2);
root.left = new Node(1);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(6);
root.right.right = new Node(8);
root.left.left.left = new Node(5);

console.log(maxNodeLevel(root));

Output

Using Depth First Search - O(n) time and O(n) space

The idea is to use depth-first traversal to visit each node while keeping track of its level, storing the count of nodes for each level in a hash map. As we traverse the tree recursively, we increment the counter for the current level in our map. After the entire tree is traversed, we simply identify which level has the highest node count.

Step by step approach:

Create a hash map to store the count of nodes for each level.
Perform DFS recursively, passing the current level information to each call.
For each node visited, increment the count for its level in the hash map.
After full traversal, iterate through the hash map to find the level with maximum nodes.
Return the level that has the highest node count.

C++

// C++ code to find Level with maximum number of nodes using DFS
#include <iostream>
#include <unordered_map>
using namespace std;

class Node {
public:
    int data;
    Node *left, *right;
    Node(int x) {
        data = x;
        left = nullptr;
        right = nullptr;
    }
};

// Helper function for DFS
void dfs(Node* root, int level, unordered_map<int, int>& levelCount) {
    if (root == nullptr)
        return;
    
    // Increment count of nodes at current level
    levelCount[level]++;
    
    // Recursively process left and right subtrees
    dfs(root->left, level + 1, levelCount);
    dfs(root->right, level + 1, levelCount);
}

int maxNodeLevel(Node *root) {
    if (root == nullptr)
        return -1;
    
    // Map to store count of nodes at each level
    unordered_map<int, int> levelCount;  
    
    // Perform DFS to count nodes at each level
    dfs(root, 0, levelCount);
    
    int maxNodes = 0;
    int maxLevel = 0;
    
    // Find the level with maximum nodes
    for (auto& pair : levelCount) {
        int level = pair.first;
        int count = pair.second;
        
        if (count > maxNodes) {
            maxNodes = count;
            maxLevel = level;
        }
    }
    
    return maxLevel;
}

int main() {
    
    //          2
    //        /   \
    //       1     3
    //      / \     \
    //     4   6     8
    //    /
    //   5
    Node* root = new Node(2);
    root->left = new Node(1);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(6);
    root->right->right = new Node(8);
    root->left->left->left = new Node(5);
    
    cout << maxNodeLevel(root) << endl;
    
    return 0;
}

Java

// Java code to find Level with maximum number of nodes using DFS
import java.util.*;

class Node {
    int data;
    Node left, right;
    Node(int x) {
        data = x;
        left = null;
        right = null;
    }
}

class GfG {

    // Helper function for DFS
    static void dfs(Node root, int level, 
    HashMap<Integer, Integer> levelCount) {
        if (root == null)
            return;

        // Increment count of nodes at current level
        levelCount.put(level, levelCount.getOrDefault(level, 0) + 1);

        // Recursively process left and right subtrees
        dfs(root.left, level + 1, levelCount);
        dfs(root.right, level + 1, levelCount);
    }

    static int maxNodeLevel(Node root) {
        if (root == null)
            return -1;

        // Map to store count of nodes at each level
        HashMap<Integer, Integer> levelCount = new HashMap<>();

        // Perform DFS to count nodes at each level
        dfs(root, 0, levelCount);

        int maxNodes = 0;
        int maxLevel = 0;

        // Find the level with maximum nodes
        for (Map.Entry<Integer, Integer> pair : levelCount.entrySet()) {
            int level = pair.getKey();
            int count = pair.getValue();

            if (count > maxNodes) {
                maxNodes = count;
                maxLevel = level;
            }
        }

        return maxLevel;
    }

    public static void main(String[] args) {

        //          2
        //        /   \
        //       1     3
        //      / \     \
        //     4   6     8
        //    /
        //   5
        Node root = new Node(2);
        root.left = new Node(1);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(6);
        root.right.right = new Node(8);
        root.left.left.left = new Node(5);

        System.out.println(maxNodeLevel(root));
    }
}

Python

# Python code to find Level with maximum number of nodes using DFS
from collections import defaultdict

class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# Helper function for DFS
def dfs(root, level, levelCount):
    if root is None:
        return

    # Increment count of nodes at current level
    levelCount[level] += 1

    # Recursively process left and right subtrees
    dfs(root.left, level + 1, levelCount)
    dfs(root.right, level + 1, levelCount)

def maxNodeLevel(root):
    if root is None:
        return -1

    # Map to store count of nodes at each level
    levelCount = defaultdict(int)

    # Perform DFS to count nodes at each level
    dfs(root, 0, levelCount)

    maxNodes = 0
    maxLevel = 0

    # Find the level with maximum nodes
    for level in levelCount:
        count = levelCount[level]
        if count > maxNodes:
            maxNodes = count
            maxLevel = level

    return maxLevel

if __name__ == "__main__":

    #          2
    #        /   \
    #       1     3
    #      / \     \
    #     4   6     8
    #    /
    #   5
    root = Node(2)
    root.left = Node(1)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(6)
    root.right.right = Node(8)
    root.left.left.left = Node(5)

    print(maxNodeLevel(root))

// C# code to find Level with maximum number of nodes using DFS
using System;
using System.Collections.Generic;

class Node {
    public int data;
    public Node left, right;
    public Node(int x) {
        data = x;
        left = null;
        right = null;
    }
}

class GfG {

    // Helper function for DFS
    static void dfs(Node root, int level, Dictionary<int, int> levelCount) {
        if (root == null)
            return;

        // Increment count of nodes at current level
        if (!levelCount.ContainsKey(level))
            levelCount[level] = 0;
        levelCount[level]++;

        // Recursively process left and right subtrees
        dfs(root.left, level + 1, levelCount);
        dfs(root.right, level + 1, levelCount);
    }

    static int maxNodeLevel(Node root) {
        if (root == null)
            return -1;

        // Map to store count of nodes at each level
        Dictionary<int, int> levelCount = new Dictionary<int, int>();

        // Perform DFS to count nodes at each level
        dfs(root, 0, levelCount);

        int maxNodes = 0;
        int maxLevel = 0;

        // Find the level with maximum nodes
        foreach (var pair in levelCount) {
            int level = pair.Key;
            int count = pair.Value;

            if (count > maxNodes) {
                maxNodes = count;
                maxLevel = level;
            }
        }

        return maxLevel;
    }

    static void Main(string[] args) {

        //          2
        //        /   \
        //       1     3
        //      / \     \
        //     4   6     8
        //    /
        //   5
        Node root = new Node(2);
        root.left = new Node(1);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(6);
        root.right.right = new Node(8);
        root.left.left.left = new Node(5);

        Console.WriteLine(maxNodeLevel(root));
    }
}

JavaScript

// JavaScript code to find Level with maximum number of nodes using DFS
class Node {
    constructor(x) {
        this.data = x;
        this.left = null;
        this.right = null;
    }
}

// Helper function for DFS
function dfs(root, level, levelCount) {
    if (root === null)
        return;

    // Increment count of nodes at current level
    if (!levelCount.has(level)) {
        levelCount.set(level, 0);
    }
    levelCount.set(level, levelCount.get(level) + 1);

    // Recursively process left and right subtrees
    dfs(root.left, level + 1, levelCount);
    dfs(root.right, level + 1, levelCount);
}

function maxNodeLevel(root) {
    if (root === null)
        return -1;

    // Map to store count of nodes at each level
    const levelCount = new Map();

    // Perform DFS to count nodes at each level
    dfs(root, 0, levelCount);

    let maxNodes = 0;
    let maxLevel = 0;

    // Find the level with maximum nodes
    for (let [level, count] of levelCount.entries()) {
        if (count > maxNodes) {
            maxNodes = count;
            maxLevel = level;
        }
    }

    return maxLevel;
}

//          2
//        /   \
//       1     3
//      / \     \
//     4   6     8
//    /
//   5
let root = new Node(2);
root.left = new Node(1);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(6);
root.right.right = new Node(8);
root.left.left.left = new Node(5);

console.log(maxNodeLevel(root));

Output

Find if there is a path between two vertices in a directed graph

kartik

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Practice Tags :

Tree

Level with maximum number of nodes

Using Breadth First Search - O(n) time and O(n) space

Using Depth First Search - O(n) time and O(n) space

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