Find Maximum Level Sum in Binary Tree using Recursion

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

Input:
      4
    /   \
   2     -5
  / \    / \
-1   3 -2   6
Output: 6
Sum of all nodes of the 1st level is 4.
Sum of all nodes of the 2nd level is -3.
Sum of all nodes of the 3rd level is 6.
Hence, the maximum sum is 6.

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

Approach: Find the maximum level in the given binary tree then create an array sum[] where sum[i] will store the sum of the elements at level i.
Now, write a recursive function that takes a node of the tree and its level as the argument and updates the sum for the current level then makes recursive calls for the children with the updated level as one more than the current level (this is because children are at a level one more than their parent). Finally, print the maximum value from the sum[] array.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// A binary tree node has data, pointer to
// the left child and the right child
struct Node {
    int data;
    struct Node *left, *right;
};
 
// 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);
}
 
// Function to return the maximum
// levels in the given tree
int maxLevel(struct Node* root)
{
    if (root == NULL)
        return 0;
    return (1 + max(maxLevel(root->left),
                    maxLevel(root->right)));
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
void maxLevelSum(struct Node* root, int max_level,
                 int sum[], int current)
{
    // Base case
    if (root == NULL)
        return;
 
    // Add current node's data to
    // its level's sum
    sum[current] += root->data;
 
    // Recursive call for the left child
    maxLevelSum(root->left, max_level, sum,
                current + 1);
 
    // Recursive call for the right child
    maxLevelSum(root->right, max_level, sum,
                current + 1);
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
int maxLevelSum(struct Node* root)
{
 
    // Maximum levels in the given tree
    int max_level = maxLevel(root);
 
    // To store the sum of every level
    int sum[max_level + 1] = { 0 };
 
    // Recursive function call to
    // update the sum[] array
    maxLevelSum(root, max_level, sum, 1);
 
    // To store the maximum sum for a level
    int maxSum = 0;
 
    // For every level of the tree, update
    // the maximum sum of a level so far
    for (int i = 1; i <= max_level; i++)
        maxSum = max(maxSum, sum[i]);
 
    // Return the maximum sum
    return maxSum;
}
 
// 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 << maxLevelSum(root);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// A binary tree node has data, pointer to
// the left child and the right child
static class Node
{
    int data;
    Node left, right;
};
 
// Helper function that allocates a
// new node with the given data and
// null left and right pointers
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Function to return the maximum
// levels in the given tree
static int maxLevel( Node root)
{
    if (root == null)
        return 0;
    return (1 + Math.max(maxLevel(root.left),
                     maxLevel(root.right)));
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
static void maxLevelSum(Node root, int max_level,
                        int sum[], int current)
{
    // Base case
    if (root == null)
        return;
 
    // Add current node's data to
    // its level's sum
    sum[current] += root.data;
 
    // Recursive call for the left child
    maxLevelSum(root.left, max_level, sum,
                current + 1);
 
    // Recursive call for the right child
    maxLevelSum(root.right, max_level, sum,
                current + 1);
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
static int maxLevelSum( Node root)
{
 
    // Maximum levels in the given tree
    int max_level = maxLevel(root);
 
    // To store the sum of every level
    int sum[] = new int[max_level + 1];
 
    // Recursive function call to
    // update the sum[] array
    maxLevelSum(root, max_level, sum, 1);
 
    // To store the maximum sum for a level
    int maxSum = 0;
 
    // For every level of the tree, update
    // the maximum sum of a level so far
    for (int i = 1; i <= max_level; i++)
        maxSum = Math.max(maxSum, sum[i]);
 
    // Return the maximum sum
    return maxSum;
}
 
// Driver code
public static void main(String args[])
{
    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 */
 
    System.out.println(maxLevelSum(root));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of above algorithm
 
# Utility class to create a node 
class Node: 
    def __init__(self, key): 
        self.val = key 
        self.left = self.right = None
 
# Helper function that allocates a
# new node with the given data and
# None left and right pointers
def newNode(data):
 
    node = Node(0)
    node.data = data
    node.left = node.right = None
    return (node)
 
# Function to return the maximum
# levels in the given tree
def maxLevel( root):
 
    if (root == None):
        return 0
    return (1 + max(maxLevel(root.left),
                    maxLevel(root.right)))
 
sum = []
 
# Function to find the maximum sum of a
# level in the tree using recursion
def maxLevelSum_( root, max_level , current):
 
    global sum
     
    # Base case
    if (root == None):
        return
 
    # Add current node's data to
    # its level's sum
    sum[current] += root.data
 
    # Recursive call for the left child
    maxLevelSum_(root.left, max_level, 
                current + 1)
 
    # Recursive call for the right child
    maxLevelSum_(root.right, max_level,
                current + 1)
 
# Function to find the maximum sum of a
# level in the tree using recursion
def maxLevelSum( root):
 
    global sum
     
    # Maximum levels in the given tree
    max_level = maxLevel(root)
 
    # To store the sum of every level
    i = 0
    sum = [None] * (max_level + 2)
    while(i <= max_level + 1):
        sum[i] = 0
        i = i + 1
 
    # Recursive function call to
    # update the sum[] array
    maxLevelSum_(root, max_level, 1)
 
    # To store the maximum sum for a level
    maxSum = 0
 
    # For every level of the tree, update
    # the maximum sum of a level so far
    i = 1
    while ( i <= max_level ):
        maxSum = max(maxSum, sum[i])
        i = i + 1
 
    # Return the maximum sum
    return maxSum
 
# Driver code
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 
     
 
print( maxLevelSum(root))
 
# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// A binary tree node has data, 
// pointer to the left child 
// and the right child
public class Node
{
    public int data;
    public Node left, right;
};
 
// Helper function that allocates a
// new node with the given data and
// null left and right pointers
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Function to return the maximum
// levels in the given tree
static int maxLevel( Node root)
{
    if (root == null)
        return 0;
    return (1 + Math.Max(maxLevel(root.left),
                         maxLevel(root.right)));
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
static void maxLevelSum(Node root, int max_level,
                        int []sum, int current)
{
    // Base case
    if (root == null)
        return;
 
    // Add current node's data to
    // its level's sum
    sum[current] += root.data;
 
    // Recursive call for the left child
    maxLevelSum(root.left, max_level, 
                sum, current + 1);
 
    // Recursive call for the right child
    maxLevelSum(root.right, max_level, 
                sum, current + 1);
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
static int maxLevelSum( Node root)
{
 
    // Maximum levels in the given tree
    int max_level = maxLevel(root);
 
    // To store the sum of every level
    int []sum = new int[max_level + 1];
 
    // Recursive function call to
    // update the sum[] array
    maxLevelSum(root, max_level, sum, 1);
 
    // To store the maximum sum for a level
    int maxSum = 0;
 
    // For every level of the tree, update
    // the maximum sum of a level so far
    for (int i = 1; i <= max_level; i++)
        maxSum = Math.Max(maxSum, sum[i]);
 
    // Return the maximum sum
    return maxSum;
}
 
// Driver code
public static void Main(String []args)
{
    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 */
 
    Console.WriteLine(maxLevelSum(root));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the approach
 
// A binary tree node has data, pointer to
// the left child and the right child
class Node
{
    constructor()
    {
        this.data=0;
        this.left=this.right=null;
    }
}
 
// Helper function that allocates a
// new node with the given data and
// null left and right pointers
function newNode(data)
{
    let node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
// Function to return the maximum
// levels in the given tree
function maxLevel(root)
{
    if (root == null)
        return 0;
    return (1 + Math.max(maxLevel(root.left),maxLevel(root.right)));
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
function maxLevelSum(root,max_level,sum,current)
{
    // Base case
    if (root == null)
        return;
   
    // Add current node's data to
    // its level's sum
    sum[current] += root.data;
   
    // Recursive call for the left child
    maxLevelSum(root.left, max_level, sum,
                current + 1);
   
    // Recursive call for the right child
    maxLevelSum(root.right, max_level, sum,
                current + 1);
}
 
// Function to find the maximum sum of a
// level in the tree using recursion
function _maxLevelSum(root)
{
    // Maximum levels in the given tree
    let max_level = maxLevel(root);
   
    // To store the sum of every level
    let sum = new Array(max_level + 1);
    for(let i=0;i<max_level+1;i++)
        sum[i]=0;
   
    // Recursive function call to
    // update the sum[] array
    maxLevelSum(root, max_level, sum, 1);
   
    // To store the maximum sum for a level
    let maxSum = 0;
   
    // For every level of the tree, update
    // the maximum sum of a level so far
    for (let i = 1; i <= max_level; i++)
        maxSum = Math.max(maxSum, sum[i]);
   
    // Return the maximum sum
    return maxSum;
}
 
// Driver code
let 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 */
 
document.write(_maxLevelSum(root));
 
 
// This code is contributed by avanitrachhadiya2155
</script>

Output:

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