Maximum sum of non-leaf nodes among all levels of the given binary tree

Given a Binary Tree having positive and negative nodes, the task is to find the maximum sum of non-leaf nodes among all levels of the given binary tree.

Examples:

Input:
                        4
                      /   \
                     2    -5
                    / \   
                  -1   3 
Output: 4
Sum of all non-leaf nodes at 0th level is 4.
Sum of all non-leaf nodes at 1st level is 2.
Sum of all non-leaf nodes at 2nd level is 0.
Hence maximum sum is 4

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

Approach: The idea to solve the above problem is to do the level order traversal of the tree. While doing traversal, process nodes of different levels separately. For every level being processed, compute the sum of non-leaf nodes in the level and keep track of the maximum sum.

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 left child
// and a pointer to right child
struct Node {
    int data;
    struct Node *left, *right;
};

// Function to return the maximum sum of non-leaf nodes
// at any level in tree using level order traversal
int maxNonLeafNodesSum(struct Node* root)
{
    // Base case
    if (root == NULL)
        return 0;

    // Initialize result
    int result = 0;

    // Do Level order traversal keeping track
    // of the 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 a node from queue
            Node* temp = q.front();
            q.pop();

            // Add non-leaf node's value to current sum
            if (temp->left != NULL || temp->right != NULL)
                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 sum of leaf nodes 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);
    cout << maxNonLeafNodesSum(root) << endl;

    return 0;
}

Java

// Java implementation of
// the above approach
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 return the maximum
// sum of non-leaf nodes  at any
// level in tree using level
// order traversal
static int maxNonLeafNodesSum(Node root)
{
  // Base case
  if (root == null)
    return 0;

  // Initialize result
  int result = 0;

  // Do Level order traversal keeping track
  // of the 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 a node
      // from queue
      Node temp = q.poll();

      // Add non-leaf node's
      // value to current sum
      if (temp.left != null ||
          temp.right != null)
        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 sum
    // of leaf nodes value
    result = max(sum, result);
  }
  return result;
}

static int max(int sum,
               int result)
{
  if (sum > result)
    return sum;
  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);
  System.out.println(maxNonLeafNodesSum(root));
}
}

// This code is contributed by sanjeev2552

Python3

# Python3 implementation of the approach
import queue

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

    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Function to return the maximum Sum of
# non-leaf nodes at any level in tree
# using level order traversal
def maxNonLeafNodesSum(root):

    # Base case
    if root == None:
        return 0

    # Initialize result
    result = 0

    # Do Level order traversal keeping track
    # of the number of nodes at every level
    q = queue.Queue()
    q.put(root)
    while not q.empty():

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

        # Iterate for all the nodes
        # in the queue currently
        Sum = 0
        while count:

            # Dequeue a node from queue
            temp = q.get()

            # Add non-leaf node's value to current Sum
            if temp.left != None or temp.right != None:
                Sum += temp.data

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

            count -= 1

        # Update the maximum Sum of leaf nodes 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)
    print(maxNonLeafNodesSum(root))

# This code is contributed by Rituraj Jain

C#

// C# implementation of
// the above approach
using System;
using System.Collections;

class GFG{

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

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

// Function to return the maximum
// sum of non-leaf nodes  at any
// level in tree using level
// order traversal
static int maxNonLeafNodesSum(Node root)
{

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

  // Initialize result
  int result = 0;

  // Do Level order traversal keeping track
  // of the number of nodes at every level
  Queue q = new Queue();

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

      // Add non-leaf node's
      // value to current sum
      if (temp.left != null ||
          temp.right != null)
        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 sum
    // of leaf nodes value
    result = max(sum, result);
  }
  return result;
}

static int max(int sum,
               int result)
{
  if (sum > result)
    return sum;

  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);
  Console.Write(maxNonLeafNodesSum(root));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

    // JavaScript implementation of the approach

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

    // Function to return the maximum
    // sum of non-leaf nodes  at any
    // level in tree using level
    // order traversal
    function maxNonLeafNodesSum(root)
    {
      // Base case
      if (root == null)
        return 0;

      // Initialize result
      let result = 0;

      // Do Level order traversal keeping track
      // of the 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 a node
          // from queue
          let temp = q[0];
          q.shift();

          // Add non-leaf node's
          // value to current sum
          if (temp.left != null ||
              temp.right != null)
            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 sum
        // of leaf nodes value
        result = max(sum, result);
      }
      return result;
    }

    function max(sum, result)
    {
      if (sum > result)
        return sum;
      return result;
    }

    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);
    document.write(maxNonLeafNodesSum(root));

</script>

Output:

8

Time Complexity: O(N) where N is the number of nodes in the given tree.
Auxiliary Space: O(N) due to queue data structure.