Check if a Binary Tree is an Even-Odd Tree or not

Given a Binary Tree, the task is to check if the binary tree is an Even-Odd binary tree or not. 

A Binary Tree is called an Even-Odd Tree when all the nodes which are at even levels have even values (assuming root to be at level 0) and all the nodes which are at odd levels have odd values.

Examples:

Input:

             2
            / \
           3   9
          / \   \
         4   10  6

Output: YES 
Explanation: 
Only node on level 0 (even) is 2 (even). 
Nodes present in level 1 are 3 and 9 (both odd). 
Nodes present in level 2 are 4, 10 and 6 (all even). 
Therefore, the Binary tree is an odd-even binary tree.

Input: 

             4
            / \
           3   7
          / \   \
         4   10  5

Output: NO 

Approach: Follow the steps below to solve the problem: 

1. The idea is to perform level-order traversal and check if the nodes present on even levels are even valued or not and nodes present on odd levels are odd valued or not.
2. If any node at an odd level is found to have odd value or vice-versa, then print “NO“.
3. Otherwise, after complete traversal of the tree, print “YES“.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

struct Node
{
    int val;
    Node *left, *right;
};

struct Node* newNode(int data)
{
    struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
    temp->val = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}

// Function to check if the
// tree is even-odd tree
bool isEvenOddBinaryTree(Node *root)
{
    if (root == NULL)
        return true;

    // Stores nodes of each level
    queue<Node*> q;
    q.push(root);

    // Store the current level
    // of the binary tree
    int level = 0;

    // Traverse until the
    // queue is empty
    while (!q.empty())
    {

        // Stores the number of nodes
        // present in the current level
        int size = q.size();

        for(int i = 0; i < size; i++)
        {
            Node *node = q.front();

            // Check if the level
            // is even or odd
            if (level % 2 == 0)
            {
                if (node->val % 2 == 1)
                    return false;
            }
            else if (level % 2 == 1)
            {
                if (node->val % 2 == 0)
                    return true;
            }

            // Add the nodes of the next
            // level into the queue
            if (node->left != NULL)
            {
                q.push(node->left);
            }
            if (node->right != NULL)
            {
                q.push(node->right);
            }
        }

        // Increment the level count
        level++;
    }
    return true;
}

// Driver Code
int main()
{

    // Construct a Binary Tree
    Node *root = NULL;
    root = newNode(2);
    root->left = newNode(3);
    root->right = newNode(9);
    root->left->left = newNode(4);
    root->left->right = newNode(10);
    root->right->right = newNode(6);

    // Check if the binary tree
    // is even-odd tree or not
    if (isEvenOddBinaryTree(root))
        cout << "YES";
    else
        cout << "NO";
}

// This code is contributed by ipg2016107

Java

// Java Program for the above approach
import java.util.*;

class GfG {

    // Tree node
    static class Node {
        int val;
        Node left, right;
    }

    // Function to return new tree node
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.val = data;
        temp.left = null;
        temp.right = null;
        return temp;
    }

    // Function to check if the
    // tree is even-odd tree
    public static boolean
    isEvenOddBinaryTree(Node root)
    {
        if (root == null)
            return true;

        // Stores nodes of each level
        Queue<Node> q
            = new LinkedList<>();
        q.add(root);

        // Store the current level
        // of the binary tree
        int level = 0;

        // Traverse until the
        // queue is empty
        while (!q.isEmpty()) {

            // Stores the number of nodes
            // present in the current level
            int size = q.size();

            for (int i = 0; i < size; i++) {
                Node node = q.poll();

                // Check if the level
                // is even or odd
                if (level % 2 == 0) {

                    if (node.val % 2 == 1)
                        return false;
                }
                else if (level % 2 == 1) {

                    if (node.val % 2 == 0)
                        return false;
                }

                // Add the nodes of the next
                // level into the queue
                if (node.left != null) {

                    q.add(node.left);
                }
                if (node.right != null) {

                    q.add(node.right);
                }
            }

            // Increment the level count
            level++;
        }

        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {

        // Construct a Binary Tree
        Node root = null;
        root = newNode(2);
        root.left = newNode(3);
        root.right = newNode(9);
        root.left.left = newNode(4);
        root.left.right = newNode(10);
        root.right.right = newNode(6);

        // Check if the binary tree
        // is even-odd tree or not
        if (isEvenOddBinaryTree(root)) {

            System.out.println("YES");
        }
        else {

            System.out.println("NO");
        }
    }
}

Python3

# Python3 program for the above approach

# Tree node
class Node:

    def __init__(self, data):

        self.left = None
        self.right = None
        self.val = data

# Function to return new tree node
def newNode(data):

    temp = Node(data)

    return temp

# Function to check if the
# tree is even-odd tree
def isEvenOddBinaryTree(root):

    if (root == None):
        return True

    q = []

    # Stores nodes of each level
    q.append(root)

    # Store the current level
    # of the binary tree
    level = 0

    # Traverse until the
    # queue is empty
    while (len(q) != 0):

        # Stores the number of nodes
        # present in the current level
        size = len(q)

        for i in range(size):
            node = q[0]
            q.pop(0)

            # Check if the level
            # is even or odd
            if (level % 2 == 0):

                if (node.val % 2 == 1):
                    return False

                elif (level % 2 == 1):
                    if (node.val % 2 == 0):
                        return False

                # Add the nodes of the next
                # level into the queue
                if (node.left != None):
                    q.append(node.left)

                if (node.right != None):
                    q.append(node.right)

            # Increment the level count
            level += 1

        return True

# Driver code
if __name__=="__main__":

    # Construct a Binary Tree
    root = None
    root = newNode(2)
    root.left = newNode(3)
    root.right = newNode(9)
    root.left.left = newNode(4)
    root.left.right = newNode(10)
    root.right.right = newNode(6)

    # Check if the binary tree
    # is even-odd tree or not
    if (isEvenOddBinaryTree(root)):
        print("YES")
    else:
        print("NO")

# This code is contributed by rutvik_56

C#

// C# Program for the
// above approach
using System;
using System.Collections.Generic;
class GfG{

// Tree node
class Node
{
  public int val;
  public Node left, right;
}

// Function to return new
// tree node
static Node newNode(int data)
{
  Node temp = new Node();
  temp.val = data;
  temp.left = null;
  temp.right = null;
  return temp;
}

// Function to check if the
// tree is even-odd tree
static bool isEvenOddBinaryTree(Node root)
{
  if (root == null)
    return true;

  // Stores nodes of each level
  Queue<Node> q = new Queue<Node>();
  q.Enqueue(root);

  // Store the current level
  // of the binary tree
  int level = 0;

  // Traverse until the
  // queue is empty
  while (q.Count != 0)
  {
    // Stores the number of nodes
    // present in the current level
    int size = q.Count;

    for (int i = 0; i < size; i++)
    {
      Node node = q.Dequeue();

      // Check if the level
      // is even or odd
      if (level % 2 == 0)
      {
        if (node.val % 2 == 1)
          return false;
      }
      else if (level % 2 == 1)
      {
        if (node.val % 2 == 0)
          return false;
      }

      // Add the nodes of the next
      // level into the queue
      if (node.left != null)
      {
        q.Enqueue(node.left);
      }
      if (node.right != null)
      {
        q.Enqueue(node.right);
      }
    }

    // Increment the level count
    level++;
  }

  return true;
}

// Driver Code
public static void Main(String[] args)
{
  // Construct a Binary Tree
  Node root = null;
  root = newNode(2);
  root.left = newNode(3);
  root.right = newNode(9);
  root.left.left = newNode(4);
  root.left.right = newNode(10);
  root.right.right = newNode(6);

  // Check if the binary tree
  // is even-odd tree or not
  if (isEvenOddBinaryTree(root))
  {
    Console.WriteLine("YES");
  }
  else
  {
    Console.WriteLine("NO");
  }
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program for the above approach

// Tree node
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.val = data;
    }
}

// Function to return new tree node
function newNode(data)
{
    let temp = new Node(data);
    return temp;
}

// Function to check if the
// tree is even-odd tree
function isEvenOddBinaryTree(root)
{
    if (root == null)
        return true;

    // Stores nodes of each level
    let q = [];
    q.push(root);

    // Store the current level
    // of the binary tree
    let level = 0;

    // Traverse until the
    // queue is empty
    while (q.length > 0)
    {

        // Stores the number of nodes
        // present in the current level
        let size = q.length;

        for(let i = 0; i < size; i++)
        {
            let node = q[0];
            q.shift();

            // Check if the level
            // is even or odd
            if (level % 2 == 0)
            {
                if (node.val % 2 == 1)
                    return false;
            }
            else if (level % 2 == 1)
            {
                if (node.val % 2 == 0)
                    return false;
            }

            // Add the nodes of the next
            // level into the queue
            if (node.left != null)
            {
                q.push(node.left);
            }
            if (node.right != null)
            {
                q.push(node.right);
            }
        }

        // Increment the level count
        level++;
    }
    return true;
}

// Driver code

// Construct a Binary Tree
let root = null;
root = newNode(2);
root.left = newNode(3);
root.right = newNode(9);
root.left.left = newNode(4);
root.left.right = newNode(10);
root.right.right = newNode(6);

// Check if the binary tree
// is even-odd tree or not
if (isEvenOddBinaryTree(root))
{
    document.write("YES");
}
else
{
    document.write("NO");
}

// This code is contributed by suresh07

</script>

Output:

YES

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

Method 2 (Using parent child difference)

Approach:
The idea is to check for the absolute difference between the child and parent node. 

1.  If the root node is odd, return “NO“.

2. If root node is even, then the child nodes should be odd, so difference should always come as odd. In true case return “YES“, else return “NO“.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// tree node
struct Node
{
    int data;
    Node *left, *right;
};

// returns a new
// tree Node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// Utility function to recursively traverse tree and check the diff between child nodes
bool BSTUtil(Node * root){
    if(root==NULL)
        return true;

    //if left nodes exist and absolute difference between left child and parent is divisible by 2, then return false
    if(root->left!=NULL && abs(root->data - root->left->data)%2==0)
        return false;
     //if right nodes exist and absolute difference between right child and parent is divisible by 2, then return false
    if(root->right!=NULL && abs(root->data - root->right->data)%2==0)
        return false;

    //recursively traverse left and right subtree
    return BSTUtil(root->left) && BSTUtil(root->right);
}

// Utility function to check if binary tree is even-odd binary tree
bool isEvenOddBinaryTree(Node * root){
    if(root==NULL)
        return true;

    // if root node is odd, return false
    if(root->data%2 != 0)
        return false;

    return BSTUtil(root);
}

// driver program
int main()
{
    // construct a tree
    Node* root = newNode(5);
    root->left = newNode(2);
    root->right = newNode(6);
    root->left->left = newNode(1);
    root->left->right = newNode(5);
    root->right->right = newNode(7);
    root->left->right->left = newNode(12);

    root->right->right->right = newNode(14);
    root->right->right->left = newNode(16);

    if(BSTUtil(root))
      cout<<"YES";
    else
      cout<<"NO";
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
  // tree node
  static class Node
  {
    public int data;
    public Node left, right;
    public Node(){
      data = 0;
      left = right = null;
    }
  }

  // returns a new
  // tree Node
  static Node newNode(int data)
  {
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
  }

  // Utility function to recursively traverse tree and check the diff between child nodes
  static boolean BSTUtil(Node root){
    if(root == null)
      return true;

    //if left nodes exist and absolute difference between left child and parent is divisible by 2, then return false
    if(root.left != null && Math.abs(root.data - root.left.data) % 2 == 0)
      return false;
    //if right nodes exist and absolute difference between right child and parent is divisible by 2, then return false
    if(root.right != null && Math.abs(root.data - root.right.data) % 2 == 0)
      return false;

    //recursively traverse left and right subtree
    return BSTUtil(root.left) && BSTUtil(root.right);
  }

  // Utility function to check if binary tree is even-odd binary tree
  static boolean isEvenOddBinaryTree(Node root){
    if(root == null)
      return true;

    // if root node is odd, return false
    if(root.data%2 != 0)
      return false;

    return BSTUtil(root);
  }


  // Driver Code
  public static void main(String args[])
  {

    // construct a tree
    Node root = newNode(5);
    root.left = newNode(2);
    root.right = newNode(6);
    root.left.left = newNode(1);
    root.left.right = newNode(5);
    root.right.right = newNode(7);
    root.left.right.left = newNode(12);

    root.right.right.right = newNode(14);
    root.right.right.left = newNode(16);

    if(BSTUtil(root))
      System.out.println("YES");
    else
      System.out.println("NO");
  }

}

// This code is contributed by shinjanpatra

Python3

# tree node
class Node:

    def __init__(self):
      self.data = 0
      self.left = self.right = None

# returns a new
# tree Node
def newNode(data):

    temp = Node()
    temp.data = data
    temp.left = temp.right = None
    return temp

# Utility function to recursively traverse tree
# and check the diff between child nodes
def BSTUtil(root):
    if(root == None):
        return True

    # if left nodes exist and absolute difference between
    # left child and parent is divisible by 2, then return False
    if(root.left != None and abs(root.data - root.left.data) % 2 == 0):
        return False
    # if right nodes exist and absolute difference between
    # right child and parent is divisible by 2, then return False
    if(root.right != None and abs(root.data - root.right.data) % 2 == 0):
        return False

    # recursively traverse left and right subtree
    return BSTUtil(root.left) and BSTUtil(root.right)

# Utility function to check if binary tree is even-odd binary tree
def isEvenOddBinaryTree(root):
    if(root == None):
      return True

    # if root node is odd, return False
    if(root.data%2 != 0):
      return False

    return BSTUtil(root)

# Driver Code

# construct a tree
root = newNode(5)
root.left = newNode(2)
root.right = newNode(6)
root.left.left = newNode(1)
root.left.right = newNode(5)
root.right.right = newNode(7)
root.left.right.left = newNode(12)

root.right.right.right = newNode(14)
root.right.right.left = newNode(16)

if(BSTUtil(root)):
    print("YES")
else:
    print("NO")

# This code is contributed by shinjanpatra

C#

// C# code for the above approach
using System;

class GFG
{

  // tree node
  class Node
  {
    public int data;
    public Node left, right;
    public Node()
    {
      data = 0;
      left = right = null;
    }
  }

  // returns a new
  // tree Node
  static Node newNode(int data)
  {
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
  }

  // Utility function to recursively traverse
  // tree and check the diff between child nodes
  static bool BSTUtil(Node root)
  {
    if (root == null)
      return true;

    //if left nodes exist and absolute difference
    // between left child and parent is divisible by 2, then return false
    if (root.left != null && Math.Abs(root.data - root.left.data) % 2 == 0)
      return false;

    // if right nodes exist and absolute difference
    // between right child and parent is divisible by 2, then return false
    if (root.right != null && Math.Abs(root.data - root.right.data) % 2 == 0)
      return false;

    // recursively traverse left and right subtree
    return BSTUtil(root.left) && BSTUtil(root.right);
  }

  // Utility function to check if binary tree is even-odd binary tree
  static bool isEvenOddBinaryTree(Node root)
  {
    if (root == null)
      return true;

    // if root node is odd, return false
    if (root.data % 2 != 0)
      return false;

    return BSTUtil(root);
  }

  // Driver Code
  static void Main(string[] args)
  {
    // construct a tree
    Node root = newNode(5);
    root.left = newNode(2);
    root.right = newNode(6);
    root.left.left = newNode(1);
    root.left.right = newNode(5);
    root.right.right = newNode(7);
    root.left.right.left = newNode(12);

    root.right.right.right = newNode(14);
    root.right.right.left = newNode(16);

    if (BSTUtil(root))
      Console.WriteLine("YES");
    else
      Console.WriteLine("NO");
  }
}

// This code is contributed by Potta Lokesh

Javascript

<script>

// tree node
class Node{

    constructor(){
      this.data = 0
      this.left = this.right = null
    }
}

// returns a new
// tree Node
function newNode(data){

    let temp = new Node()
    temp.data = data
    temp.left = temp.right = null
    return temp
}

// Utility function to recursively traverse tree
// and check the diff between child nodes
function BSTUtil(root){
    if(root == null)
        return true

    // if left nodes exist and absolute difference between
    // left child and parent is divisible by 2, then return false
    if(root.left != null && Math.abs(root.data - root.left.data) % 2 == 0)
        return false
    // if right nodes exist and absolute difference between
    // right child and parent is divisible by 2, then return false
    if(root.right != null && Math.abs(root.data - root.right.data) % 2 == 0)
        return false

    // recursively traverse left and right subtree
    return BSTUtil(root.left) && BSTUtil(root.right)
}

// Utility function to check if binary tree is even-odd binary tree
function isEvenOddBinaryTree(root){
    if(root == null)
      return true

    // if root node is odd, return false
    if(root.data%2 != 0)
      return false

    return BSTUtil(root)
}

// Driver Code

// construct a tree
let root = newNode(5)
root.left = newNode(2)
root.right = newNode(6)
root.left.left = newNode(1)
root.left.right = newNode(5)
root.right.right = newNode(7)
root.left.right.left = newNode(12)

root.right.right.right = newNode(14)
root.right.right.left = newNode(16)

if(BSTUtil(root))
    document.write("YES","</br>")
else
    document.write("NO","</br>")

// This code is contributed by shinjanpatra

</script>

Output

YES

Time Complexity: O(N)

Auxiliary Space: O(1)

Another Approach(Using recursion):
A simple and efficient approach as compared to above approach.
Follow the below steps to solve the given problem:
1). We simply traverse the whole binary tree recursively in inorder fashing and keep track to level at each node.
2). We will check at each level that current node data is follow the even-odd tree constraints. If the node don’t follow the constraint the we store the false in ans and simply return.
3). In driver code we will print our answer based on ans variable.

Below is the implementation of above approach:

C++

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;

// structure of tree node
struct Node{
    int data;
    Node *left, *right;
};

// returns a new tree node
Node* newNode(int data){
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

//function to check if binary tree is even-odd binary tree
void isEvenOddBinaryTree(Node * root, bool& ans, int level){
    // base case
    if(root == NULL) return;
    isEvenOddBinaryTree(root->left, ans, level+1);
    // checking for even level
    if(level % 2 == 0){
        if(root->data % 2 != 0) ans = false;
    }
    // checking for odd level
    else{
        if(root->data % 2 == 0) ans = false;
    }
    isEvenOddBinaryTree(root->right, ans, level+1);
}

// driver program to test above function
int main(){
    // Construct a Binary Tree
    Node *root = NULL;
    root = newNode(2);
    root->left = newNode(3);
    root->right = newNode(9);
    root->left->left = newNode(4);
    root->left->right = newNode(10);
    root->right->right = newNode(6);

    bool ans = true;
    isEvenOddBinaryTree(root, ans, 0);
    if(ans)
        cout<<"YES";
    else
        cout<<"NO";
    return 0;
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

Java

// Java code to implement the above approach

import java.io.*;
import java.util.*;

// structure of tree node
class Node {
    int data;
    Node left, right;
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class GFG {

    // function to check if binary tree is even-odd binary
    // tree
    static void isEvenOddBinaryTree(Node root, boolean ans,
                                    int level)
    {
        // base case
        if (root == null)
            return;
        isEvenOddBinaryTree(root.left, ans, level + 1);
        // checking for even level
        if (level % 2 == 0) {
            if (root.data % 2 != 0)
                ans = false;
        }
        // checking for odd level
        else {
            if (root.data % 2 == 0)
                ans = false;
        }
        isEvenOddBinaryTree(root.right, ans, level + 1);
    }

    public static void main(String[] args)
    {
        // Construct a Binary Tree
        Node root = null;
        root = new Node(2);
        root.left = new Node(3);
        root.right = new Node(9);
        root.left.left = new Node(4);
        root.left.right = new Node(10);
        root.right.right = new Node(6);
        boolean ans = true;
        isEvenOddBinaryTree(root, ans, 0);
        if (ans)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

// This code is contributed by karthik.

Python

# Python program for the above approach
# class of node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# returns a new node
def newNode(data):
    return Node(data)


# furction to check if binary tree is even - odd binary tree
ans = True
def isEvenOddBinaryTree(root, level):
    global ans
    # base case
    if(root is None):
        return
    isEvenOddBinaryTree(root.left, level+1)
    # check for even level
    if(level % 2 == 0):
        if(root.data % 2 != 0):
            ans = False
    else:
        if(root.data % 2 == 0):
            ans = False
    isEvenOddBinaryTree(root.right, level+1)


# driver program to test above function
# construct a binary tree
root = None
root = newNode(2)
root.left = newNode(3)
root.right = newNode(9)
root.left.left = newNode(4)
root.left.right = newNode(10)
root.right.right = newNode(6)

isEvenOddBinaryTree(root, 0)
if(ans):
    print("Yes")
else:
    print("No")

C#

// C# code to implement the above approach
using System;
using System.Collections;

class Gfg
{
  // structure of tree node
  class Node{
    public int data;
    public Node left, right;
    public Node(int data)
    {
      this.data=data;
      this.left=null;
      this.right=null;
    }
  }


  //function to check if binary tree is even-odd binary tree
  static void isEvenOddBinaryTree(Node root, bool ans, int level)
  {
    // base case
    if(root == null)
      return;
    isEvenOddBinaryTree(root.left, ans, level+1);
    // checking for even level
    if(level % 2 == 0){
      if(root.data % 2 != 0)
        ans = false;
    }
    // checking for odd level
    else{
      if(root.data % 2 == 0)
        ans = false;
    }
    isEvenOddBinaryTree(root.right, ans, level+1);
  }

  // driver program to test above function
  static void Main(string[] args)
  {
    // Construct a Binary Tree
    Node root = null;
    root = new Node(2);
    root.left = new Node(3);
    root.right = new Node(9);
    root.left.left = new Node(4);
    root.left.right = new Node(10);
    root.right.right = new Node(6);

    bool ans = true;
    isEvenOddBinaryTree(root, ans, 0);
    if(ans)
      Console.Write("YES");
    else
      Console.Write("NO");
  }
}

// This code is contributed by poojaagarwal2.

Javascript

// JavaScript Program for the above approach
// structure of node
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// returns a new node
function newNode(data){
    return new Node(data);
}

// function to check if binary tree is even - odd binary tree
let ans = true;
function isEvenOddBinaryTree(root, level){
    // base case
    if(root == null) return;
    isEvenOddBinaryTree(root.left, level+1);
    // check for even level
    if(level % 2 == 0){
        if(root.data % 2 != 0) ans = false;
    }
    // checking for odd level
    else{
        if(root.data % 2 == 0) ans = false;
    }
    isEvenOddBinaryTree(root.right, level+1);
}

// driver program to test above function
// construct a binary tree
let root = null;
root = newNode(2);
root.left = newNode(3);
root.right = newNode(9);
root.left.left = newNode(4);
root.left.right = newNode(10);
root.right.right = newNode(6);

isEvenOddBinaryTree(root, 0);
if(ans)
    console.log("YES");
else
    console.log("NO");

// This code is contributed by KIRTI AGARWAL(KIRTIAGARWAL23121999)

Output

YES

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of given binary tree.