Largest value in each level of Binary Tree

Difficulty Level : Medium
Last Updated : 03 Nov, 2020

Given a binary tree, find the largest value in each level.

Examples :

Input :
        1
       / \
      2   3 
Output : 1 3

Input : 
        4
       / \
      9   2
     / \   \
    3   5   7 
Output : 4 9 7

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : The idea is to recursively traverse tree in a pre-order fashion. Root is considered to be at zeroth level. While traversing, keep track of the level of the element and if its current level is not equal to the number of elements present in the list, update the maximum element at that level in the list.

Below is the implementation to find largest value on each level of Binary Tree.

C++

// C++ program to find largest
// value on each level of binary tree.
#include <bits/stdc++.h>
using namespace std;
  
/* A binary tree node has data,
pointer to left child and a 
pointer to right child */
struct Node {
    int val;
    struct Node *left, *right;
};
  
/* Recursive function to find
the largest value on each level */
void helper(vector<int>& res, Node* root, int d)
{
    if (!root)
        return;
  
    // Expand list size
    if (d == res.size())
        res.push_back(root->val);
  
    else
  
        // to ensure largest value 
        // on level is being stored
        res[d] = max(res[d], root->val);
  
    // Recursively traverse left and
    // right subtrees in order to find
    // out the largest value on each level
    helper(res, root->left, d + 1);
    helper(res, root->right, d + 1);
}
  
// function to find largest values
vector<int> largestValues(Node* root)
{
    vector<int> res;
    helper(res, root, 0);
    return res;
}
  
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->val = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Driver code
int main()
{
    /* Let us construct a Binary Tree
        4
       / \
      9   2
     / \   \
    3   5   7 */
  
    Node* root = NULL;
    root = newNode(4);
    root->left = newNode(9);
    root->right = newNode(2);
    root->left->left = newNode(3);
    root->left->right = newNode(5);
    root->right->right = newNode(7);
      
    vector<int> res = largestValues(root);
    for (int i = 0; i < res.size(); i++)
        cout << res[i] << " ";
          
    return 0;
}

Java

// Java program to find largest 
// value on each level of binary tree.
import java.util.*;
  
class GFG 
{
  
/* A binary tree node has data, 
pointer to left child and a 
pointer to right child */
static class Node 
{ 
    int val; 
    Node left, right; 
}; 
  
/* Recursive function to find 
the largest value on each level */
static void helper(Vector<Integer> res, Node root, int d) 
{ 
    if (root == null) 
        return; 
  
    // Expand list size 
    if (d == res.size()) 
        res.add(root.val); 
  
    else
  
        // to ensure largest value 
        // on level is being stored 
        res.set(d, Math.max(res.get(d), root.val)); 
  
    // Recursively traverse left and 
    // right subtrees in order to find 
    // out the largest value on each level 
    helper(res, root.left, d + 1); 
    helper(res, root.right, d + 1); 
} 
  
// function to find largest values 
static Vector<Integer> largestValues(Node root) 
{ 
    Vector<Integer> res = new Vector<>(); 
    helper(res, root, 0); 
    return res; 
} 
  
/* Helper function that allocates a 
new node with the given data and 
NULL left and right pointers. */
static Node newNode(int data) 
{ 
    Node temp = new Node(); 
    temp.val = data; 
    temp.left = temp.right = null; 
    return temp; 
} 
  
// Driver code 
public static void main(String[] args) 
{
    /* Let us construct a Binary Tree 
        4 
    / \ 
    9 2 
    / \ \ 
    3 5 7 */
  
    Node root = null; 
    root = newNode(4); 
    root.left = newNode(9); 
    root.right = newNode(2); 
    root.left.left = newNode(3); 
    root.left.right = newNode(5); 
    root.right.right = newNode(7); 
      
    Vector<Integer> res = largestValues(root); 
    for (int i = 0; i < res.size(); i++) 
            System.out.print(res.get(i)+" ");
}
}
  
/* This code is contributed by PrinciRaj1992 */

Python3

# Python program to find largest value
# on each level of binary tree.
  
""" Recursive function to find 
the largest value on each level """
def helper(res, root, d): 
  
    if ( not root): 
        return
  
    # Expand list size 
    if (d == len(res)): 
        res.append(root.val) 
  
    else:
  
        # to ensure largest value 
        # on level is being stored 
        res[d] = max(res[d], root.val) 
  
    # Recursively traverse left and 
    # right subtrees in order to find 
    # out the largest value on each level 
    helper(res, root.left, d + 1) 
    helper(res, root.right, d + 1) 
  
  
# function to find largest values 
def largestValues(root): 
  
    res = [] 
    helper(res, root, 0) 
    return res
  
  
# Helper function that allocates a new 
# node with the given data and None left 
# and right pointers.                                     
class newNode: 
  
    # Constructor to create a new node 
    def __init__(self, data): 
        self.val = data 
        self.left = None
        self.right = None
  
          
# Driver Code 
if __name__ == '__main__':
    """ Let us construct the following Tree
        4 
        / \ 
        9 2 
    / \ \
    3 5 7 """
    root = newNode(4) 
    root.left = newNode(9) 
    root.right = newNode(2) 
    root.left.left = newNode(3)
    root.left.right = newNode(5)
    root.right.right = newNode(7)
    print(*largestValues(root))                         
  
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to find largest 
// value on each level of binary tree.
using System;
using System.Collections.Generic;
  
class GFG 
{
  
/* A binary tree node has data, 
pointer to left child and a 
pointer to right child */
public class Node 
{ 
    public int val; 
    public Node left, right; 
}; 
  
/* Recursive function to find 
the largest value on each level */
static void helper(List<int> res, 
                   Node root, int d) 
{ 
    if (root == null) 
        return; 
  
    // Expand list size 
    if (d == res.Count) 
        res.Add(root.val); 
  
    else
  
        // to ensure largest value 
        // on level is being stored 
        res[d] = Math.Max(res[d], root.val); 
  
    // Recursively traverse left and 
    // right subtrees in order to find 
    // out the largest value on each level 
    helper(res, root.left, d + 1); 
    helper(res, root.right, d + 1); 
} 
  
// function to find largest values 
static List<int> largestValues(Node root) 
{ 
    List<int> res = new List<int>(); 
    helper(res, root, 0); 
    return res; 
} 
  
/* Helper function that allocates a 
new node with the given data and 
NULL left and right pointers. */
static Node newNode(int data) 
{ 
    Node temp = new Node(); 
    temp.val = data; 
    temp.left = temp.right = null; 
    return temp; 
} 
  
// Driver code 
public static void Main(String[] args) 
{
      
    /* Let us construct a Binary Tree 
        4 
    / \ 
    9 2 
    / \ \ 
    3 5 7 */
    Node root = null; 
    root = newNode(4); 
    root.left = newNode(9); 
    root.right = newNode(2); 
    root.left.left = newNode(3); 
    root.left.right = newNode(5); 
    root.right.right = newNode(7); 
      
    List<int> res = largestValues(root); 
    for (int i = 0; i < res.Count; i++) 
        Console.Write(res[i] + " ");
}
}
  
// This code is contributed by 29AjayKumar