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