Find maximum level sum in Binary Tree - GeeksforGeeks

11 May, 2021
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++
Java
Python3
C#
Javascript
// 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 traversalint 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 codeint 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;}
// A queue based Java program to find maximum// sum of a level in Binary Treeimport java.util.LinkedList;import java.util.Queue; class GFG{ // A binary tree node has data, pointer// to left child and a pointer to right// childstatic 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 traversalstatic 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 codepublic 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
# A queue based Python3 program to find# maximum sum of a level in Binary Treefrom collections import deque # A binary tree node has data, pointer# to left child and a pointer to right# childclass 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 traversaldef 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 codeif __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
// A queue based C# program to find maximum// sum of a level in Binary Treeusing 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
<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 codelet 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>
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.
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