Maximum absolute difference between any two level sum in a Binary Tree
Given a Binary Tree having positive and negative nodes, the task is to find the maximum absolute difference of level sum in it.
Examples:
Input: 4 / \ 2 -5 / \ / \ -1 3 -2 6 Output: 9 Explanation: Sum of all nodes of 0 level is 4 Sum of all nodes of 1 level is -3 Sum of all nodes of 2 level is 6 Hence maximum absolute difference of level sum = 9 (6 - (-3)) Input: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 Output: 16
Approach: To find the maximum absolute difference of level sum, we only need to find Maximum level sum and Minimum level sum because the absolute difference of maximum and minimum level sum always gives us Maximum absolute difference, i.e.
Maximum absolute difference = abs(Maximum level sum – Minimum level sum)
Below are the steps for the algorithm of the above observation:
- The idea is to do 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 maximum and minimum level sum.
- Then return the absolute difference of maximum and minimum level sum.
Below is the implementation of the above approach:
C++
// C++ program to find the maximum // absolute difference of level // sum in Binary Tree #include <bits/stdc++.h> using namespace std; // Class containing left and // right child of current // node and key value struct Node { int data; Node *left, *right; }; Node *newNode( int data) { Node *node = new Node(); node->data = data; node->left = NULL; node->right = NULL; return node; } // Function to find the maximum // absolute difference of level // sum in binary tree // using level order traversal int maxAbsDiffLevelSum(Node *root) { // Initialize value of maximum // and minimum level sum int maxsum = INT_MIN; int minsum = INT_MAX; queue<Node *> qu; qu.push(root); // Do Level order traversal // keeping track of number // of nodes at every level. while (!qu.empty()) { // Get the size of queue when // the level order traversal // for one level finishes int sz = qu.size(); // Iterate for all the nodes in // the queue currently int sum = 0; for ( int i = 0; i < sz; i++) { // Dequeue an node from queue Node *t = qu.front(); qu.pop(); // Add this node's value to // the current sum. sum += t->data; // Enqueue left and // right children of // dequeued node if (t->left != NULL) qu.push(t->left); if (t->right != NULL) qu.push(t->right); } // Update the maximum // level sum value maxsum = max(maxsum, sum); // Update the minimum // level sum value minsum = min(minsum, sum); } // return the maximum absolute // difference of level sum return abs (maxsum - minsum); } // Driver code int main() { Node *root = new Node(); root = newNode(4); root->left = newNode(2); root->right = newNode(-5); root->left->left = newNode(-1); root->left->right = newNode(3); root->right->left = newNode(-2); root->right->right = newNode(6); /* Constructed Binary tree is: 4 / \ 2 -5 / \ / \ -1 3 -2 6 */ cout << maxAbsDiffLevelSum(root) << endl; } // This code is contributed by sanjeev2552 |
Java
// Java program to find the maximum // absolute difference of level // sum in Binary Tree import java.util.*; // Class containing left and // right child of current // node and key value class Node { int data; Node left, right; public Node( int item) { data = item; left = right = null ; } } class BinaryTree { // Root of the Binary Tree Node root; public BinaryTree() { root = null ; } // Function to find // the maximum absolute // difference of level // sum in binary tree // using level order traversal public int maxAbsDiffLevelSum() { // Initialize value of maximum // and minimum level sum int maxsum = Integer.MIN_VALUE; int minsum = Integer.MAX_VALUE; Queue<Node> qu = new LinkedList<>(); qu.offer(root); // Do Level order traversal // keeping track of number // of nodes at every level. while (!qu.isEmpty()) { // Get the size of queue when // the level order traversal // for one level finishes int sz = qu.size(); // Iterate for all the nodes in // the queue currently int sum = 0 ; for ( int i = 0 ; i < sz; i++) { // Dequeue an node from queue Node t = qu.poll(); // Add this node's value to // the current sum. sum += t.data; // Enqueue left and // right children of // dequeued node if (t.left != null ) qu.offer(t.left); if (t.right != null ) qu.offer(t.right); } // Update the maximum // level sum value maxsum = Math.max(maxsum, sum); // Update the minimum // level sum value minsum = Math.min(minsum, sum); } // return the maximum absolute // difference of level sum return Math.abs(maxsum - minsum); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node( 4 ); tree.root.left = new Node( 2 ); tree.root.right = new Node(- 5 ); tree.root.left.left = new Node(- 1 ); tree.root.left.right = new Node( 3 ); tree.root.right.left = new Node(- 2 ); tree.root.right.right = new Node( 6 ); /* Constructed Binary tree is: 4 / \ 2 -5 / \ / \ -1 3 -2 6 */ System.out.println( tree.maxAbsDiffLevelSum()); } } |
Python3
# Python 3 program to find # the maximum absolute difference # of level sum in Binary Tree import sys # Class containing left and # right child of current # node and key value class newNode: def __init__( self , data): self .data = data self .left = None self .right = None # Function to find the maximum # absolute difference of level # sum in binary tree # using level order traversal def maxAbsDiffLevelSum(root): # Initialize value of maximum # and minimum level sum maxsum = - sys.maxsize - 1 minsum = sys.maxsize qu = [] qu.append(root) # Do Level order traversal # keeping track of number # of nodes at every level. while ( len (qu) > 0 ): # Get the size of queue when # the level order traversal # for one level finishes sz = len (qu) # Iterate for all the nodes in # the queue currently sum = 0 for i in range (sz): # Dequeue an node from # queue t = qu[ 0 ] qu.remove(qu[ 0 ]) # Add this node's value # to the current sum. sum + = t.data # Enqueue left and # right children of # dequeued node if (t.left ! = None ): qu.append(t.left) if (t.right ! = None ): qu.append(t.right) # Update the maximum # level sum value maxsum = max (maxsum, sum ) # Update the minimum # level sum value minsum = min (minsum, sum ) # return the maximum absolute # difference of level sum return abs (maxsum - minsum) # Driver code if __name__ = = '__main__' : root = newNode( 4 ) root.left = newNode( 2 ) root.right = newNode( - 5 ) root.left.left = newNode( - 1 ) root.left.right = newNode( 3 ) root.right.left = newNode( - 2 ) root.right.right = newNode( 6 ) '''/* Constructed Binary tree is: 4 / \ 2 -5 / / / \ -1 3 -2 6 */ ''' print (maxAbsDiffLevelSum(root)) # This code is contributed by SURENDRA_GANGWAR |
C#
// C# program to find the maximum // absolute difference of level // sum in Binary Tree using System; using System.Collections.Generic; // Class to represent Tree node public class Node { public int data; public Node left, right; public Node( int item) { data = item; left = null ; right = null ; } } class BinaryTree{ Node root; // Function to find the maximum // absolute difference of level // sum in binary tree // using level order traversal public int maxAbsDiffLevelSum() { // Initialize value of maximum // and minimum level sum int maxsum = Int32.MinValue; int minsum = Int32.MaxValue; Queue<Node> qu = new Queue<Node>(); qu.Enqueue(root); // Do Level order traversal // keeping track of number // of nodes at every level. while (qu.Count != 0) { // Get the size of queue when // the level order traversal // for one level finishes int sz = qu.Count; // Iterate for all the nodes in // the queue currently int sum = 0; for ( int i = 0; i < sz; i++) { // Dequeue an node from queue Node t = qu.Dequeue(); // Add this node's value to // the current sum. sum += t.data; // Enqueue left and // right children of // dequeued node if (t.left != null ) qu.Enqueue(t.left); if (t.right != null ) qu.Enqueue(t.right); } // Update the maximum // level sum value maxsum = Math.Max(maxsum, sum); // Update the minimum // level sum value minsum = Math.Min(minsum, sum); } // Return the maximum absolute // difference of level sum return Math.Abs(maxsum - minsum); } // Driver code static public void Main () { BinaryTree tree = new BinaryTree(); tree.root = new Node(4); tree.root.left = new Node(2); tree.root.right = new Node(-5); tree.root.left.left = new Node(-1); tree.root.left.right = new Node(3); tree.root.right.left = new Node(-2); tree.root.right.right = new Node(6); /* Constructed Binary tree is: 4 / \ 2 -5 / \ / \ -1 3 -2 6 */ Console.WriteLine(tree.maxAbsDiffLevelSum()); } } // This code is contributed by offbeat |
Javascript
<script> // Javascript program to find the maximum // absolute difference of level // sum in Binary Tree // Class to represent Tree node class Node { constructor(item) { this .left = null ; this .right = null ; this .data = item; } } let root; // Function to find the maximum // absolute difference of level // sum in binary tree // using level order traversal function maxAbsDiffLevelSum() { // Initialize value of maximum // and minimum level sum let maxsum = Number.MIN_VALUE; let minsum = Number.MAX_VALUE; let qu = []; qu.push(root); // Do Level order traversal // keeping track of number // of nodes at every level. while (qu.length != 0) { // Get the size of queue when // the level order traversal // for one level finishes let sz = qu.length; // Iterate for all the nodes in // the queue currently let sum = 0; for (let i = 0; i < sz; i++) { // Dequeue an node from queue let t = qu.shift(); // Add this node's value to // the current sum. sum += t.data; // Enqueue left and // right children of // dequeued node if (t.left != null ) qu.push(t.left); if (t.right != null ) qu.push(t.right); } // Update the maximum // level sum value maxsum = Math.max(maxsum, sum); // Update the minimum // level sum value minsum = Math.min(minsum, sum); } // Return the maximum absolute // difference of level sum return Math.abs(maxsum - minsum); } root = new Node(4); root.left = new Node(2); root.right = new Node(-5); root.left.left = new Node(-1); root.left.right = new Node(3); root.right.left = new Node(-2); root.right.right = new Node(6); /* Constructed Binary tree is: 4 / \ 2 -5 / \ / \ -1 3 -2 6 */ document.write(maxAbsDiffLevelSum()); // This code is contributed by divyeshrabadiya07. </script> |
Output:
9
Time Complexity: O(N)
Auxiliary Space: O(N)
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