Find the maximum node at a given level in a binary tree

Given a Binary Tree and a Level. The task is to find the node with the maximum value at that given level.

The idea is to traverse the tree along depth recursively and return the nodes once the required level is reached and then return the maximum of left and right subtrees for each subsequent call. So that the last call will return the node with maximum value among all nodes at the given level.



Below is the step by step algorithm:

  1. Perform DFS traversal and every time decrease the value of level by 1 and keep traversing to the left and right subtrees recursively.
  2. When value of level becomes 0, it means we are on the given level, then return root->data.
  3. Find the maximum between the two values returned by left and right subtrees and return the maximum.

Below is the implementation of above approach:

C++

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// CPP program to find the node with
// maximum value at a given level
  
#include <iostream>
  
using namespace std;
  
// Tree node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// Utility function to create a new Node
struct Node* newNode(int val)
{
    struct Node* temp = new Node;
    temp->left = NULL;
    temp->right = NULL;
    temp->data = val;
    return temp;
}
  
// function to find the maximum value
// at given level
int maxAtLevel(struct Node* root, int level)
{
    // If the tree is empty
    if (root == NULL)
        return 0;
  
    // if level becomes 0, it means we are on
    // any node at the given level
    if (level == 0)
        return root->data;
  
    int x = maxAtLevel(root->left, level - 1);
    int y = maxAtLevel(root->right, level - 1);
  
    // return maximum of two
    return max(x, y);
}
  
// Driver code
int main()
{
    // Creating the tree
    struct Node* root = NULL;
    root = newNode(45);
    root->left = newNode(46);
    root->left->left = newNode(18);
    root->left->left->left = newNode(16);
    root->left->left->right = newNode(23);
    root->left->right = newNode(17);
    root->left->right->left = newNode(24);
    root->left->right->right = newNode(21);
    root->right = newNode(15);
    root->right->left = newNode(22);
    root->right->left->left = newNode(37);
    root->right->left->right = newNode(41);
    root->right->right = newNode(19);
    root->right->right->left = newNode(49);
    root->right->right->right = newNode(29);
  
    int level = 3;
  
    cout << maxAtLevel(root, level);
  
    return 0;
}

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Java














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// Java program to find the  


// node with maximum value  


// at a given level 


import java.util.*; 


class GFG 


{ 


  


// Tree node 


static class Node  


{ 


    int data; 


    Node left, right; 


} 


  


// Utility function to 


// create a new Node 


static Node newNode(int val) 


{ 


    Node temp = new Node(); 


    temp.left = null; 


    temp.right = null; 


    temp.data = val; 


    return temp; 


} 


  


// function to find  


// the maximum value 


// at given level 


static int maxAtLevel(Node root, int level) 


{ 


    // If the tree is empty 


    if (root == null) 


        return 0; 


  


    // if level becomes 0,  


    // it means we are on 


    // any node at the given level 


    if (level == 0) 


        return root.data; 


  


    int x = maxAtLevel(root.left, level - 1); 


    int y = maxAtLevel(root.right, level - 1); 


  


    // return maximum of two 


    return Math.max(x, y); 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    // Creating the tree 


    Node root = null; 


    root = newNode(45); 


    root.left = newNode(46); 


    root.left.left = newNode(18); 


    root.left.left.left = newNode(16); 


    root.left.left.right = newNode(23); 


    root.left.right = newNode(17); 


    root.left.right.left = newNode(24); 


    root.left.right.right = newNode(21); 


    root.right = newNode(15); 


    root.right.left = newNode(22); 


    root.right.left.left = newNode(37); 


    root.right.left.right = newNode(41); 


    root.right.right = newNode(19); 


    root.right.right.left = newNode(49); 


    root.right.right.right = newNode(29); 


  


    int level = 3; 


  


    System.out.println(maxAtLevel(root, level)); 


} 


} 


  


// This code is contributed 


// by Arnab Kundu 



















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Python3














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# Python3 program to find the node   


# with maximum value at a given level  


  


# Helper function that allocates a new  


# node with the given data and None  


# left and right poers.                                      


class newNode:  


  


    # Constructor to create a new node  


    def __init__(self, data):  


        self.data = data 


        self.left = None


        self.right = None


  


# function to find the maximum   


# value at given level  


def maxAtLevel(root, level):  


  


    # If the tree is empty  


    if (root == None) : 


        return 0


  


    # if level becomes 0, it means we  


    # are on any node at the given level  


    if (level == 0) : 


        return root.data  


  


    x = maxAtLevel(root.left, level - 1)  


    y = maxAtLevel(root.right, level - 1)  


  


    # return maximum of two  


    return max(x, y) 


      


# Driver Code  


if __name__ == '__main__': 


  


    """  


    Let us create Binary Tree shown 


    in above example """


    root = newNode(45)  


    root.left = newNode(46)  


    root.left.left = newNode(18)  


    root.left.left.left = newNode(16)  


    root.left.left.right = newNode(23)  


    root.left.right = newNode(17)  


    root.left.right.left = newNode(24)  


    root.left.right.right = newNode(21)  


    root.right = newNode(15)  


    root.right.left = newNode(22)  


    root.right.left.left = newNode(37)  


    root.right.left.right = newNode(41)  


    root.right.right = newNode(19)  


    root.right.right.left = newNode(49)  


    root.right.right.right = newNode(29) 


    level = 3


    print(maxAtLevel(root, level)) 


  


# This code is contributed by 


# Shubham Singh(SHUBHAMSINGH10) 



















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C#














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// C# program to find the  


// node with maximum value  


// at a given level 


using System; 


  


class GFG 


{ 


  


    // Tree node 


    class Node  


    { 


        public int data; 


        public Node left, right; 


    } 


  


    // Utility function to 


    // create a new Node 


    static Node newNode(int val) 


    { 


        Node temp = new Node(); 


        temp.left = null; 


        temp.right = null; 


        temp.data = val; 


        return temp; 


    } 


  


    // function to find  


    // the maximum value 


    // at given level 


    static int maxAtLevel(Node root, int level) 


    { 


        // If the tree is empty 


        if (root == null) 


            return 0; 


  


        // if level becomes 0,  


        // it means we are on 


        // any node at the given level 


        if (level == 0) 


            return root.data; 


  


        int x = maxAtLevel(root.left, level - 1); 


        int y = maxAtLevel(root.right, level - 1); 


  


        // return maximum of two 


        return Math.Max(x, y); 


    } 


  


    // Driver code 


    public static void Main(String []args) 


    { 


        // Creating the tree 


        Node root = null; 


        root = newNode(45); 


        root.left = newNode(46); 


        root.left.left = newNode(18); 


        root.left.left.left = newNode(16); 


        root.left.left.right = newNode(23); 


        root.left.right = newNode(17); 


        root.left.right.left = newNode(24); 


        root.left.right.right = newNode(21); 


        root.right = newNode(15); 


        root.right.left = newNode(22); 


        root.right.left.left = newNode(37); 


        root.right.left.right = newNode(41); 


        root.right.right = newNode(19); 


        root.right.right.left = newNode(49); 


        root.right.right.right = newNode(29); 


  


        int level = 3; 


  


        Console.WriteLine(maxAtLevel(root, level)); 


    } 


} 


  


// This code is contributed by 29AjayKumar 



















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Output:


49








    

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