Find maximum level sum in Binary Tree

Given a Binary Tree having positive and negative nodes, the task is to find the maximum sum level in it.

Examples:

Input :               4
                    /   \
                   2    -5
                  / \    /\
                -1   3 -2  6
Output: 6
Explanation :
Sum of all nodes of 0'th level is 4
Sum of all nodes of 1'th level is -3
Sum of all nodes of 0'th level is 6
Hence maximum sum is 6

Input :          1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7  
Output :  17

This problem is a variation of the maximum width problem. The idea is to do a level order traversal of the tree. While doing traversal, process nodes of different levels separately. For every level being processed, compute the sum of nodes in the level and keep track of the maximum sum.

Below is the implementation of the above idea:

C++

// A queue based C++ program to find maximum sum
// of a level in 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 data;
    struct Node *left, *right;
};

// Function to find the maximum sum of a level in tree
// using level order traversal
int maxLevelSum(struct Node* root)
{
    // Base case
    if (root == NULL)
        return 0;

    // Initialize result
    int result = root->data;

    // Do Level order traversal keeping track of number
    // of nodes at every level.
    queue<Node*> q;
    q.push(root);
    while (!q.empty())
    {
        // Get the size of queue when the level order
        // traversal for one level finishes
        int count = q.size();

        // Iterate for all the nodes in the queue currently
        int sum = 0;
        while (count--)
        {
            // Dequeue an node from queue
            Node* temp = q.front();
            q.pop();

            // Add this node's value to current sum.
            sum = sum + temp->data;

            // Enqueue left and right children of
            // dequeued node
            if (temp->left != NULL)
                q.push(temp->left);
            if (temp->right != NULL)
                q.push(temp->right);
        }

        // Update the maximum node count value
        result = max(sum, result);
    }

    return result;
}

/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}

// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(6);
    root->right->right->right = newNode(7);

    /*   Constructed Binary tree is:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7    */
    cout << "Maximum level sum is " << maxLevelSum(root)
         << endl;
    return 0;
}

Java

// A queue based Java program to find maximum
// sum of a level in Binary Tree
import java.util.LinkedList;
import java.util.Queue;

class GFG{

// A binary tree node has data, pointer
// to left child and a pointer to right
// child
static class Node
{
    int data;
    Node left, right;

    public Node(int data)
    {
        this.data = data;
        this.left = this.right = null;
    }
};

// Function to find the maximum
// sum of a level in tree
// using level order traversal
static int maxLevelSum(Node root)
{

    // Base case
    if (root == null)
        return 0;

    // Initialize result
    int result = root.data;

    // Do Level order traversal keeping
    // track of number of nodes at every
    // level.
    Queue<Node> q = new LinkedList<>();
    q.add(root);
    while (!q.isEmpty())
    {

        // Get the size of queue when the
        // level order traversal for one
        // level finishes
        int count = q.size();

        // Iterate for all the nodes
        // in the queue currently
        int sum = 0;
        while (count-- > 0)
        {

            // Dequeue an node from queue
            Node temp = q.poll();

            // Add this node's value
            // to current sum.
            sum = sum + temp.data;

            // Enqueue left and right children
            // of dequeued node
            if (temp.left != null)
                q.add(temp.left);
            if (temp.right != null)
                q.add(temp.right);
        }

        // Update the maximum node
        // count value
        result = Math.max(sum, result);
    }
    return result;
}

// Driver code
public static void main(String[] args)
{
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.right = new Node(8);
    root.right.right.left = new Node(6);
    root.right.right.right = new Node(7);

    /*   Constructed Binary tree is:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7    */
    System.out.println("Maximum level sum is " +
                        maxLevelSum(root));
}
}

// This code is contributed by sanjeev2552

Python3

# A queue based Python3 program to find
# maximum sum of a level in Binary Tree
from collections import deque

# A binary tree node has data, pointer
# to left child and a pointer to right
# child
class Node:

    def __init__(self, key):

        self.data = key
        self.left = None
        self.right = None

# Function to find the maximum sum
# of a level in tree
# using level order traversal
def maxLevelSum(root):

    # Base case
    if (root == None):
        return 0

    # Initialize result
    result = root.data

    # Do Level order traversal keeping
    # track of number
    # of nodes at every level.
    q = deque()
    q.append(root)

    while (len(q) > 0):

        # Get the size of queue when the
        # level order traversal for one
        # level finishes
        count = len(q)

        # Iterate for all the nodes in
        # the queue currently
        sum = 0
        while (count > 0):

            # Dequeue an node from queue
            temp = q.popleft()

            # Add this node's value to current sum.
            sum = sum + temp.data

            # Enqueue left and right children of
            # dequeued node
            if (temp.left != None):
                q.append(temp.left)
            if (temp.right != None):
                q.append(temp.right)

            count -= 1

        # Update the maximum node count value
        result = max(sum, result)

    return result

# Driver code
if __name__ == '__main__':

    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.right = Node(8)
    root.right.right.left = Node(6)
    root.right.right.right = Node(7)

    # Constructed Binary tree is:
    #              1
    #            /   \
    #          2      3
    #        /  \      \
    #       4    5      8
    #                 /   \
    #                6     7
    print("Maximum level sum is", maxLevelSum(root))

# This code is contributed by mohit kumar 29

C#

// A queue based C# program to find maximum
// sum of a level in 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 data;
      public
        Node left, right;

      public Node(int data)
      {
        this.data = data;
        this.left = this.right = null;
      }
    };

  // Function to find the maximum
  // sum of a level in tree
  // using level order traversal
  static int maxLevelSum(Node root)
  {

    // Base case
    if (root == null)
      return 0;

    // Initialize result
    int result = root.data;

    // Do Level order traversal keeping
    // track of number of nodes at every
    // level.
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
    while (q.Count != 0)
    {

      // Get the size of queue when the
      // level order traversal for one
      // level finishes
      int count = q.Count;

      // Iterate for all the nodes
      // in the queue currently
      int sum = 0;
      while (count --> 0)
      {

        // Dequeue an node from queue
        Node temp = q.Dequeue();

        // Add this node's value
        // to current sum.
        sum = sum + temp.data;

        // Enqueue left and right children
        // of dequeued node
        if (temp.left != null)
          q.Enqueue(temp.left);
        if (temp.right != null)
          q.Enqueue(temp.right);
      }

      // Update the maximum node
      // count value
      result = Math.Max(sum, result);
    }
    return result;
  }

  // Driver code
  public static void Main(String[] args)
  {
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.right = new Node(8);
    root.right.right.left = new Node(6);
    root.right.right.right = new Node(7);

    /*   Constructed Binary tree is:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7    */
    Console.WriteLine("Maximum level sum is " +
                      maxLevelSum(root));
  }
}

// This code is contributed by gauravrajput1

Javascript

<script>
// A queue based Javascript program to find maximum
// sum of a level in Binary Tree


    // A binary tree node has data, pointer
// to left child and a pointer to right
// child
    class Node
    {
        constructor(data)
        {
            this.data = data;
            this.left = this.right = null;
        }
    }

// Function to find the maximum
// sum of a level in tree
// using level order traversal
function maxLevelSum(root)
{
    // Base case
    if (root == null)
        return 0;

    // Initialize result
    let result = root.data;

    // Do Level order traversal keeping
    // track of number of nodes at every
    // level.
    let q = [];
    q.push(root);
    while (q.length!=0)
    {

        // Get the size of queue when the
        // level order traversal for one
        // level finishes
        let count = q.length;

        // Iterate for all the nodes
        // in the queue currently
        let sum = 0;
        while (count-- > 0)
        {

            // Dequeue an node from queue
            let temp = q.shift();

            // Add this node's value
            // to current sum.
            sum = sum + temp.data;

            // Enqueue left and right children
            // of dequeued node
            if (temp.left != null)
                q.push(temp.left);
            if (temp.right != null)
                q.push(temp.right);
        }

        // Update the maximum node
        // count value
        result = Math.max(sum, result);
    }
    return result;
}


// Driver code
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(8);
root.right.right.left = new Node(6);
root.right.right.right = new Node(7);

 /*   Constructed Binary tree is:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7    */

document.write("Maximum level sum is " +
                        maxLevelSum(root));


    // This code is contributed by unknown2108
</script>

Output

Maximum level sum is 17

Complexity Analysis:

Time Complexity: O(N) where N is the total number of nodes in the tree.
In level order traversal, every node of the tree is processed once, and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the sum at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N).

Auxiliary Space: O(w) where w is the maximum width of the tree.
In level order traversal, a queue is maintained whose maximum size at any moment can go up to the maximum width of the binary tree.

Using Recursion Without Queue:-

• We will use recursion and do any dfs of the tree.
• As in dfs we will move downwards of the tree so while moving we will take care of the level of the tree
• We will add the node value to the current level of the tree
• In the end we will return the maximum sum from all level

Implementation:-

• We will start traversal by level 0 that is from root
• From root we will move downwards using recursion and while moving we will increase the level of the tree by +1.
• We will take a unordered_map to store the sum of the current level
• In the end we will return the maximum value from the map.

C++

// A queue based C++ program to find maximum sum
// of a level in 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 data;
    struct Node *left, *right;
};

//function to get sum or each level
void dfs(Node* root,int level,unordered_map<int,int> &mm)
{
  //base condition
  if(!root)return;

  //adding root value to its level sum
  mm[level]+=root->data;

  //increasing level
  level++;

  //moving left
  dfs(root->left,level,mm);

  //moving right
  dfs(root->right,level,mm);
}
// Function to find the maximum sum of a level in tree
// using level order traversal
int maxLevelSum(struct Node* root)
{
    // Base case
    if (root == NULL)
        return 0;

      //map to store sum of each level
      unordered_map<int,int> mm;

      //calling function
      dfs(root,0,mm);

      //variable to store answer
    int result = INT_MIN;

      //iterating over map
      for(auto x:mm)result = max(x.second,result);

    return result;
}

/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}

// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(6);
    root->right->right->right = newNode(7);

    /*   Constructed Binary tree is:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7    */
    cout << "Maximum level sum is " << maxLevelSum(root)
         << endl;
    return 0;
}

//code contributed by shubhamrajput6156

Java

import java.util.*;

// A binary tree node has data, a pointer to the left child,
// and a pointer to the right child
class Node {
    int data;
    Node left, right;

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

class GFG {
    // Function to get the sum for each level
    static void dfs(Node root, int level, Map<Integer, Integer> mm) {
        // Base condition
        if (root == null) return;

        // Adding the root value to its level sum
        mm.put(level, mm.getOrDefault(level, 0) + root.data);

        // Increasing level
        level++;

        // Moving left
        dfs(root.left, level, mm);

        // Moving right
        dfs(root.right, level, mm);
    }

    // Function to find the maximum sum of a level in
  // the tree using level order traversal
    static int maxLevelSum(Node root) {
        // Base case
        if (root == null) return 0;

        // Map to store the sum of each level
        Map<Integer, Integer> mm = new HashMap<>();

        // Calling the dfs function
        dfs(root, 0, mm);

        // Variable to store the answer
        int result = Integer.MIN_VALUE;

        // Iterating over the map
        for (Map.Entry<Integer, Integer> entry : mm.entrySet()) {
            result = Math.max(entry.getValue(), result);
        }

        return result;
    }

    // Driver code
    public static void main(String[] args) {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(8);
        root.right.right.left = new Node(6);
        root.right.right.right = new Node(7);

        // Constructed binary tree is:
        //        1
        //      /   \
        //     2     3
        //    / \     \
        //   4   5     8
        //          /   \
        //         6     7

        System.out.println("Maximum level sum is " + maxLevelSum(root));
    }
}

Output:- Maximum level sum is 17

Time Complexity:- O(N)

Space Complexity:- O(H) where H is height of tree 

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.