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++ 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 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());
}
} |
# 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# 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 |
<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> |
9
Time Complexity: O(N)
Auxiliary Space: O(N)