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Create loops of even and odd values in a binary tree
  • Difficulty Level : Easy
  • Last Updated : 08 Feb, 2021

Given a binary tree with the node structure containing a data part, left and right pointers and an arbitrary pointer(abtr). The node’s value can be any positive integer. The problem is to create odd and even loops in a binary tree. An odd loop is a loop which connects all the nodes having odd numbers and similarly even loop is for nodes having even numbers. To create such loops, the abtr pointer of each node is used. An abtr pointer of an odd node(node having odd number) points to some other odd node in the tree. A loop must be created in such way that from any node we could traverse all the nodes in the loop to which the node belongs.
Examples: 
 

Consider the binary tree given below 

       1             
    /     \          
   2        3            
 /   \     /  \       
4     5   6    7   
     
Now with the help of abtr pointers of node, 
we connect odd and even nodes as:

Odd loop
1 -> 5 -> 3 -> 7 -> 1(again pointing to first node
                      in the loop)               
    
Even loop
2 -> 4 -> 6 -> 2(again pointing to first node
                 in the loop)

Nodes in the respective loop can be arranged in
any order. But from any node in the loop we should 
be able to traverse all the nodes in the loop.

 

Approach: The following steps are: 
 

  1. Add pointers of the nodes having even and odd numbers to even_ptrs and odd_ptrs arrays respectively. Through any tree traversal we could get the respective node pointers.
  2. For both the even_ptrs and odd_ptrs array, perform:
    • As the array contains node pointers, consider an element at ith index, let it be node, and the assign node->abtr = element at (i+1)th index.
    • For last element of the array, node->abtr = element at index 0.

 

CPP




// C++ implementation to create odd and even loops
// in a binary tree
#include <bits/stdc++.h>
 
using namespace std;
 
// structure of a node
struct Node
{
    int data;
    Node *left, *right, *abtr;
};
 
// Utility function to create a new node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = node->abtr = NULL;
    return node;
}
 
// preorder traversal to place the node pointer
// in the respective even_ptrs or odd_ptrs list
void preorderTraversal(Node *root, vector<Node*> *even_ptrs,
                       vector<Node*> *odd_ptrs)
{
    if (!root)
        return;
     
    // place node ptr in even_ptrs list if
    // node contains even number 
    if (root->data % 2 == 0)   
        (*even_ptrs).push_back(root);
         
    // else place node ptr in odd_ptrs list
    else
        (*odd_ptrs).push_back(root);
         
    preorderTraversal(root->left, even_ptrs, odd_ptrs);
    preorderTraversal(root->right, even_ptrs, odd_ptrs);
}
 
// function to create the even and odd loops
void createLoops(Node *root)
{
     
    vector<Node*> even_ptrs, odd_ptrs;
    preorderTraversal(root, &even_ptrs, &odd_ptrs);
     
    int i;
     
    // forming even loop
    for (i=1; i<even_ptrs.size(); i++)
        even_ptrs[i-1]->abtr = even_ptrs[i];
         
    // for the last element
    even_ptrs[i-1]->abtr = even_ptrs[0];   
     
    // Similarly forming odd loop
    for (i=1; i<odd_ptrs.size(); i++)
        odd_ptrs[i-1]->abtr = odd_ptrs[i];
    odd_ptrs[i-1]->abtr = odd_ptrs[0];
}
 
// traversing the loop from any random
// node in the loop
void traverseLoop(Node *start)
{
    Node *curr = start;
    do
    {
        cout << curr->data << " ";
        curr = curr->abtr;
    } while (curr != start);
}
 
// Driver program to test above
int main()
{
    // Binary tree formation
    struct Node* root = NULL;
    root = newNode(1);                   /*          1          */
    root->left = newNode(2);             /*       /    \        */
    root->right = newNode(3);            /*      2       3      */
    root->left->left = newNode(4);       /*    /  \    /   \    */
    root->left->right = newNode(5);      /*   4    5  6     7   */
    root->right->left = newNode(6);
    root->right->right = newNode(7);
     
    createLoops(root);
     
    // traversing odd loop from any
    // random odd node
    cout << "Odd nodes: ";
    traverseLoop(root->right);
     
    cout << endl << "Even nodes: ";
    // traversing even loop from any
    // random even node
    traverseLoop(root->left);   
     
    return 0;
}


Java




// Java implementation to create odd and even loops
// in a binary tree
import java.util.*;
class GFG
{
 
// structure of a node
static class Node
{
    int data;
    Node left, right, abtr;
};
 
// Utility function to create a new node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = node.abtr = null;
    return node;
}
static Vector<Node> even_ptrs = new Vector<>();
static Vector<Node> odd_ptrs = new Vector<>();
 
// preorder traversal to place the node pointer
// in the respective even_ptrs or odd_ptrs list
static void preorderTraversal(Node root)
{
    if (root == null)
        return;
     
    // place node ptr in even_ptrs list if
    // node contains even number 
    if (root.data % 2 == 0)   
        (even_ptrs).add(root);
         
    // else place node ptr in odd_ptrs list
    else
        (odd_ptrs).add(root);
         
    preorderTraversal(root.left);
    preorderTraversal(root.right);
}
 
// function to create the even and odd loops
static void createLoops(Node root)
{
    preorderTraversal(root);  
    int i;
     
    // forming even loop
    for (i = 1; i < even_ptrs.size(); i++)
        even_ptrs.get(i - 1).abtr = even_ptrs.get(i);
         
    // for the last element
    even_ptrs.get(i - 1).abtr = even_ptrs.get(0);   
     
    // Similarly forming odd loop
    for (i = 1; i < odd_ptrs.size(); i++)
        odd_ptrs.get(i - 1).abtr = odd_ptrs.get(i);
    odd_ptrs.get(i - 1).abtr = odd_ptrs.get(0);
}
 
// traversing the loop from any random
// node in the loop
static void traverseLoop(Node start)
{
    Node curr = start;
    do
    {
        System.out.print(curr.data + " ");
        curr = curr.abtr;
    } while (curr != start);
}
 
// Driver code
public static void main(String[] args)
{
   
    // Binary tree formation
    Node root = null;
    root = newNode(1);                   /*          1          */
    root.left = newNode(2);             /*       /    \        */
    root.right = newNode(3);            /*      2       3      */
    root.left.left = newNode(4);       /*    /  \    /   \    */
    root.left.right = newNode(5);      /*   4    5  6     7   */
    root.right.left = newNode(6);
    root.right.right = newNode(7);  
    createLoops(root);
     
    // traversing odd loop from any
    // random odd node
    System.out.print("Odd nodes: ");
    traverseLoop(root.right);  
    System.out.print( "\nEven nodes: ");
   
    // traversing even loop from any
    // random even node
    traverseLoop(root.left);   
}
}
 
// This code is contributed by aashish1995


Python3




# Python3 implementation to create odd and even loops
# in a binary tree
 
# structure of a node
class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None
        self.abtr = None
 
even_ptrs  = []
odd_ptrs = []
 
# preorder traversal to place the node pointer
# in the respective even_ptrs or odd_ptrs list
def preorderTraversal(root):
    global even_ptrs, odd_ptrs
 
    if (not root):
        return
 
    # place node ptr in even_ptrs list if
    # node contains even number
    if (root.data % 2 == 0):
        even_ptrs.append(root)
         
    # else place node ptr in odd_ptrs list
    else:
        odd_ptrs.append(root)
 
    preorderTraversal(root.left)
    preorderTraversal(root.right)
 
# function to create the even and odd loops
def createLoops(root):
    preorderTraversal(root)
 
    # forming even loop
    i = 1
    while i < len(even_ptrs):
        even_ptrs[i - 1].abtr = even_ptrs[i]
        i += 1
 
    # for the last element
    even_ptrs[i - 1].abtr = even_ptrs[0]
 
    # Similarly forming odd loop
    i = 1
    while i < len(odd_ptrs):
        odd_ptrs[i - 1].abtr = odd_ptrs[i]
        i += 1
    odd_ptrs[i - 1].abtr = odd_ptrs[0]
 
#traversing the loop from any random
#node in the loop
def traverseLoop(start):
    curr = start
 
    while True and curr:
        print(curr.data, end = " ")
        curr = curr.abtr
 
        if curr == start:
            break
    print()
 
# Driver program to test above
if __name__ == '__main__':
   
    # Binary tree formation
    root = None
    root = Node(1)                 #/*         1         */
    root.left = Node(2)        #     /*     / \     */
    root.right = Node(3)         #/*     2     3     */
    root.left.left = Node(4)     #/* / \ / \ */
    root.left.right = Node(5)     #/* 4 5 6     7 */
    root.right.left = Node(6)
    root.right.right = Node(7)
 
    createLoops(root)
 
    # traversing odd loop from any
    # random odd node
    print("Odd nodes:", end = " ")
    traverseLoop(root.right)
 
    print("Even nodes:", end = " ")
     
    # traversing even loop from any
    # random even node
    traverseLoop(root.left)
 
# This code is contributed by mohit kumar 29


C#




// C# implementation to create odd and even loops
// in a binary tree
using System;
using System.Collections.Generic;
class GFG
{
 
  // structure of a node
  public
    class Node
    {
      public
        int data;
      public
        Node left, right, abtr;
    };
 
  // Utility function to create a new node
  static Node newNode(int data)
  {
    Node node = new Node();
    node.data = data;
    node.left = node.right = node.abtr = null;
    return node;
  }
  static List<Node> even_ptrs = new List<Node>();
  static List<Node> odd_ptrs = new List<Node>();
 
  // preorder traversal to place the node pointer
  // in the respective even_ptrs or odd_ptrs list
  static void preorderTraversal(Node root)
  {
    if (root == null)
      return;
 
    // place node ptr in even_ptrs list if
    // node contains even number 
    if (root.data % 2 == 0)   
      (even_ptrs).Add(root);
 
    // else place node ptr in odd_ptrs list
    else
      (odd_ptrs).Add(root);
    preorderTraversal(root.left);
    preorderTraversal(root.right);
  }
 
  // function to create the even and odd loops
  static void createLoops(Node root)
  {
    preorderTraversal(root);  
    int i;
 
    // forming even loop
    for (i = 1; i < even_ptrs.Count; i++)
      even_ptrs[i - 1].abtr = even_ptrs[i];
 
    // for the last element
    even_ptrs[i - 1].abtr = even_ptrs[0];   
 
    // Similarly forming odd loop
    for (i = 1; i < odd_ptrs.Count; i++)
      odd_ptrs[i - 1].abtr = odd_ptrs[i];
    odd_ptrs[i - 1].abtr = odd_ptrs[0];
  }
 
  // traversing the loop from any random
  // node in the loop
  static void traverseLoop(Node start)
  {
    Node curr = start;
    do
    {
      Console.Write(curr.data + " ");
      curr = curr.abtr;
    } while (curr != start);
  }
 
  // Driver code
  public static void Main(String[] args)
  {
 
    // Binary tree formation
    Node root = null;
    root = newNode(1);                   /*          1          */
    root.left = newNode(2);             /*       /    \        */
    root.right = newNode(3);            /*      2       3      */
    root.left.left = newNode(4);       /*    /  \    /   \    */
    root.left.right = newNode(5);      /*   4    5  6     7   */
    root.right.left = newNode(6);
    root.right.right = newNode(7);  
    createLoops(root);
 
    // traversing odd loop from any
    // random odd node
    Console.Write("Odd nodes: ");
    traverseLoop(root.right);  
    Console.Write( "\nEven nodes: ");
 
    // traversing even loop from any
    // random even node
    traverseLoop(root.left);   
  }
}
 
 
// This code is contributed by gauravrajput1


Output: 
 

Odd nodes: 3 7 1 5
Even nodes: 2 4 6

Time Complexity: Equal to the time complexity of any recursive tree traversal which is O(n)
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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