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

Below is the implementation of the above idea:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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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 upto the maximum width of the binary 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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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