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

Last Updated : 27 Jan, 2023

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 0^th level is 4. Sum of all non-leaf nodes at 1^st level is 2. Sum of all non-leaf nodes at 2^nd 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# 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>