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 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 algorithm of above observation:



  1. The idea is to do level order traversal of the tree.
  2. While doing traversal, process nodes of different levels separately.
  3. For every level being processed, compute sum of nodes in the level and keep track of maximum and minimum level sum.
  4. Then return the absolute difference of maximum and minimum level sum.

Below is the implementation of the above approach:

Java

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// 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());
    }
}

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Output:

9

Time Complexity: O(N)
Auxiliary Space: O(N)

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