Given a binary tree, the task is to find the absolute difference between the even valued and the odd valued nodes in a binary tree.
Examples:
Input:
5
/ \
2 6
/ \ \
1 4 8
/ / \
3 7 9
Output: 5
Explanation:
Sum of the odd value nodes is:
5 + 1 + 3 + 7 + 9 = 25
Sum of the even value nodes is:
2 + 6 + 4 + 8 = 20
Absolute difference = (25 – 20) = 5.
Input:
4
/ \
1 4
/ \ \
7 2 6
Output: 8
Approach:
Follow the steps below to solve the problem:
- Traverse each node in the tree and check if the value at that node is odd or even.
- Update oddSum and evenSum accordingly after visiting each node.
- After complete traversal of the tree, print the absolute difference between oddSum and evenSum.
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std;int oddsum = 0;int evensum = 0;int ans = 0;struct node { int data; struct node* left; struct node* right;};struct node* newnode(int data){ node* temp = new node(); temp->data = data; temp->left = temp->right = NULL; return temp;}// Function calculate the sum of// odd and even value nodevoid OddEvenDifference(struct node* root){ // If root is NULL if (root == NULL) { return; } else { // Check if current root // is odd or even if (root->data % 2 == 0) { evensum += root->data; } else { oddsum += root->data; } // Call on the left subtree OddEvenDifference(root->left); // Call on the right subtree OddEvenDifference(root->right); }}// Driver Codeint main(){ node* root = newnode(5); root->left = newnode(2); root->right = newnode(6); root->left->left = newnode(1); root->left->right = newnode(4); root->left->right->left = newnode(3); root->right->right = newnode(8); root->right->right->right = newnode(9); root->right->right->left = newnode(7); OddEvenDifference(root); cout << abs(oddsum - evensum) << endl;} |
Java
// Java implementation of// the above approachclass GFG{static int oddsum = 0;static int evensum = 0;static int ans = 0;static class node{ int data; node left; node right;};static node newnode(int data){ node temp = new node(); temp.data = data; temp.left = temp.right = null; return temp;}// Function calculate the sum of// odd and even value nodestatic void OddEvenDifference(node root){ // If root is null if (root == null) { return; } else { // Check if current root // is odd or even if (root.data % 2 == 0) { evensum += root.data; } else { oddsum += root.data; } // Call on the left subtree OddEvenDifference(root.left); // Call on the right subtree OddEvenDifference(root.right); }}// Driver Codepublic static void main(String[] args){ node root = newnode(5); root.left = newnode(2); root.right = newnode(6); root.left.left = newnode(1); root.left.right = newnode(4); root.left.right.left = newnode(3); root.right.right = newnode(8); root.right.right.right = newnode(9); root.right.right.left = newnode(7); OddEvenDifference(root); System.out.print(Math.abs( oddsum - evensum) + "\n");}}// This code is contributed by amal kumar choubey |
Python3
# Python3 implementation of# the above approachoddsum = 0evensum = 0class Node: def __init__(self, data): self.left = None self.right = None self.data = data def newnode(data): temp = Node(data) return temp# Function calculate the sum of# odd and even value nodedef OddEvenDifference(root): global evensum, oddsum # If root is NULL if (root == None): return else: # Check if current root # is odd or even if (root.data % 2 == 0): evensum += root.data else: oddsum += root.data # Call on the left subtree OddEvenDifference(root.left) # Call on the right subtree OddEvenDifference(root.right) # Driver codeif __name__=="__main__": root = newnode(5) root.left = newnode(2) root.right = newnode(6) root.left.left = newnode(1) root.left.right = newnode(4) root.left.right.left= newnode(3) root.right.right = newnode(8) root.right.right.right= newnode(9) root.right.right.left= newnode(7) OddEvenDifference(root) print(abs(oddsum - evensum)) # This code is contributed by rutvik_56 |
C#
// C# implementation of// the above approachusing System;class GFG{static int oddsum = 0;static int evensum = 0;//static int ans = 0;class node{ public int data; public node left; public node right;};static node newnode(int data){ node temp = new node(); temp.data = data; temp.left = temp.right = null; return temp;}// Function calculate the sum of// odd and even value nodestatic void OddEvenDifference(node root){ // If root is null if (root == null) { return; } else { // Check if current root // is odd or even if (root.data % 2 == 0) { evensum += root.data; } else { oddsum += root.data; } // Call on the left subtree OddEvenDifference(root.left); // Call on the right subtree OddEvenDifference(root.right); }}// Driver Codepublic static void Main(String[] args){ node root = newnode(5); root.left = newnode(2); root.right = newnode(6); root.left.left = newnode(1); root.left.right = newnode(4); root.left.right.left = newnode(3); root.right.right = newnode(8); root.right.right.right = newnode(9); root.right.right.left = newnode(7); OddEvenDifference(root); Console.Write(Math.Abs( oddsum - evensum) + "\n");}}// This code is contributed by amal kumar choubey |
Javascript
<script> // Javascript implementation of the above approach let oddsum = 0; let evensum = 0; let ans = 0; class node { constructor(data) { this.data = data; this.left = this.right = null; } } function newnode(data) { let temp = new node(data); return temp; } // Function calculate the sum of // odd and even value node function OddEvenDifference(root) { // If root is null if (root == null) { return; } else { // Check if current root // is odd or even if (root.data % 2 == 0) { evensum += root.data; } else { oddsum += root.data; } // Call on the left subtree OddEvenDifference(root.left); // Call on the right subtree OddEvenDifference(root.right); } } let root = newnode(5); root.left = newnode(2); root.right = newnode(6); root.left.left = newnode(1); root.left.right = newnode(4); root.left.right.left = newnode(3); root.right.right = newnode(8); root.right.right.right = newnode(9); root.right.right.left = newnode(7); OddEvenDifference(root); document.write(Math.abs(oddsum - evensum) + "</br>");</script> |
Output:
5
Time Complexity: O(N) where N is the number of nodes in given Binary Tree, as it does a simple traversal of the tree.
Auxiliary Space: O(h) where h is the height of given Binary Tree due to Recursion.
Another Approach (Iterative):
We can solve this problem by level order traversal using queue.
Below is the implementation of above approach:
C++
// C++ Program for the above approach#include<bits/stdc++.h>using namespace std;// structure of nodestruct Node{ int data; struct Node* left = NULL; struct Node* right = NULL; Node(int data){ this->data = data; this->left = this->right = NULL; }};// Utility function to create nodeNode* newnode(int data){ return new Node(data);}// function to find the absolute differenceint OddEvenDifference(Node* root){ int oddSum = 0; int evenSum = 0; queue<Node*> q; q.push(root); while(!q.empty()){ Node* front_node = q.front(); q.pop(); if(front_node->data % 2 == 0) evenSum += front_node->data; else oddSum += front_node->data; if(front_node->left != NULL) q.push(front_node->left); if(front_node->right != NULL) q.push(front_node->right); } return abs(oddSum - evenSum);}// Driver codeint main(){ Node* root = newnode(5); root->left = newnode(2); root->right = newnode(6); root->left->left = newnode(1); root->left->right = newnode(4); root->left->right->left = newnode(3); root->right->right = newnode(8); root->right->right->right = newnode(9); root->right->right->left = newnode(7); cout<<OddEvenDifference(root); return 0;}// This code is contributed by Kirti Agarwal(kirtiagarwal23121999) |
Python3
# Python Program for the above approachimport queue# structure of nodeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None# function to find the absolute differencedef OddEvenDifference(root): oddSum = 0 evenSum = 0 # create a queue q = queue.Queue() q.put(root) while(not q.empty()): front_node = q.get() # get the front node from the queue if(front_node.data % 2 == 0): # if the data of the front node is even evenSum += front_node.data # add it to the even sum else: oddSum += front_node.data # otherwise, add it to the odd sum if(front_node.left is not None): # if the front node has a left child q.put(front_node.left) # add it to the queue if(front_node.right is not None): # if the front node has a right child q.put(front_node.right) # add it to the queue return abs(oddSum - evenSum) # return the absolute difference between the odd and even sums# Driver codeif __name__ == '__main__': root = Node(5) root.left = Node(2) root.right = Node(6) root.left.left = Node(1) root.left.right = Node(4) root.left.right.left = Node(3) root.right.right = Node(8) root.right.right.right = Node(9) root.right.right.left = Node(7) # call the OddEvenDifference function and print the result print(OddEvenDifference(root)) |
Java
// Java Program for the above approachimport java.util.*;// structure of nodeclass Node{ int data; Node left; Node right; Node(int data){ this.data = data; this.left = null; this.right = null; }};// Driver codeclass Main { // function to find the absolute difference static int OddEvenDifference(Node root){ int oddSum = 0; int evenSum = 0; Queue<Node> q = new LinkedList<Node>(); q.add(root); while(!q.isEmpty()){ Node front_node = q.poll(); if(front_node.data % 2 == 0) evenSum += front_node.data; else oddSum += front_node.data; if(front_node.left != null) q.add(front_node.left); if(front_node.right != null) q.add(front_node.right); } return Math.abs(oddSum - evenSum); } public static void main(String[] args) { Node root = new Node(5); root.left = new Node(2); root.right = new Node(6); root.left.left = new Node(1); root.left.right = new Node(4); root.left.right.left = new Node(3); root.right.right = new Node(8); root.right.right.right = new Node(9); root.right.right.left = new Node(7); System.out.println(OddEvenDifference(root)); }} |
Javascript
// JavaScript program for the above approach// class of tree nodeclass Node{ constructor(data){ this.data = data; this.left = null; this.right = null; }}// utility function to create nodefunction newnode(data){ return new Node(data);}// function to find the absolute differencefunction oddEvenDifference(root){ let oddSum = 0; let evenSum = 0; let q = []; q.push(root); while(q.length > 0){ let front_node = q.shift(); if(front_node.data % 2 == 0) evenSum += front_node.data; else oddSum += front_node.data; if(front_node.left) q.push(front_node.left); if(front_node.right) q.push(front_node.right); } return Math.abs(oddSum - evenSum);}// driver codelet root = newnode(5);root.left = newnode(2);root.right = newnode(6);root.left.left = newnode(1);root.left.right = newnode(4);root.left.right.left = newnode(3);root.right.right = newnode(8);root.right.right.right = newnode(9);root.right.right.left = newnode(7);console.log(oddEvenDifference(root)); |
C#
// C# Program for the above approachusing System;using System.Collections.Generic;// structure of nodepublic class Node { public int data; public Node left = null; public Node right = null; public Node(int data) { this.data = data; this.left = this.right = null; }}class MainClass { // Utility function to create node static Node newnode(int data) { return new Node(data); } // function to find the absolute difference static int OddEvenDifference(Node root) { int oddSum = 0; int evenSum = 0; Queue<Node> q = new Queue<Node>(); q.Enqueue(root); while (q.Count > 0) { Node front_node = q.Peek(); q.Dequeue(); if (front_node.data % 2 == 0) evenSum += front_node.data; else oddSum += front_node.data; if (front_node.left != null) q.Enqueue(front_node.left); if (front_node.right != null) q.Enqueue(front_node.right); } return Math.Abs(oddSum - evenSum); } // Driver code static void Main(string[] args) { Node root = newnode(5); root.left = newnode(2); root.right = newnode(6); root.left.left = newnode(1); root.left.right = newnode(4); root.left.right.left = newnode(3); root.right.right = newnode(8); root.right.right.right = newnode(9); root.right.right.left = newnode(7); Console.WriteLine(OddEvenDifference(root)); }}// This code is contributed by user_dtewbxkn77n |
Output
5
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