Create loops of even and odd values in a binary tree - GeeksforGeeks

13 May, 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 that 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 an odd number) points to some other odd node in the tree. A loop must be created in such a 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:  
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.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 assigned node->abtr = element at (i+1)th index.For last element of the array, node->abtr = element at index 0.
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.
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 assigned node->abtr = element at (i+1)th index.For last element of the array, node->abtr = element at index 0.
As the array contains node pointers, consider an element at ith index, let it be node, and the assigned node->abtr = element at (i+1)th index.
For last element of the array, node->abtr = element at index 0.
 
CPP
Java
Python3
C#
Javascript
// C++ implementation to create odd and even loops// in a binary tree#include <bits/stdc++.h> using namespace std; // structure of a nodestruct Node{    int data;    Node *left, *right, *abtr;}; // Utility function to create a new nodestruct 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 listvoid 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 loopsvoid 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 loopvoid traverseLoop(Node *start){    Node *curr = start;    do    {        cout << curr->data << " ";        curr = curr->abtr;    } while (curr != start);} // Driver program to test aboveint 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 implementation to create odd and even loops// in a binary treeimport java.util.*;class GFG{ // structure of a nodestatic class Node{    int data;    Node left, right, abtr;}; // Utility function to create a new nodestatic 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 liststatic 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 loopsstatic 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 loopstatic void traverseLoop(Node start){    Node curr = start;    do    {        System.out.print(curr.data + " ");        curr = curr.abtr;    } while (curr != start);} // Driver codepublic 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 implementation to create odd and even loops# in a binary tree # structure of a nodeclass 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 listdef 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 loopsdef 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 loopdef traverseLoop(start):    curr = start     while True and curr:        print(curr.data, end = " ")        curr = curr.abtr         if curr == start:            break    print() # Driver program to test aboveif __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# implementation to create odd and even loops// in a binary treeusing 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
<script> // Javascript implementation to// create odd and even loops// in a binary tree // structure of a nodeclass Node{    // Utility function to create a new node    constructor(data)    {        this.data=data;        this.left=this.right=this.abtr =null;    }     } let even_ptrs = [];let odd_ptrs = []; // preorder traversal to place the node pointer// in the respective even_ptrs or odd_ptrs listfunction preorderTraversal(root){    if (root == null)        return;          // place node ptr in even_ptrs list if    // node contains even number    if (root.data % 2 == 0)          (even_ptrs).push(root);              // else place node ptr in odd_ptrs list    else        (odd_ptrs).push(root);              preorderTraversal(root.left);    preorderTraversal(root.right);} // function to create the even and odd loopsfunction createLoops(root){    preorderTraversal(root);     let i;          // forming even loop    for (i = 1; i < even_ptrs.length; 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.length; 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 loopfunction traverseLoop(start){    let curr = start;    do    {        document.write(curr.data + " ");        curr = curr.abtr;    } while (curr != start);} // Driver code         /*          1          */        /*       /    \        */        /*      2       3      */        /*    /  \    /   \    */        /*   4    5  6     7   */   // Binary tree formationlet root = null;root = new Node(1);                  root.left = new Node(2);            root.right = new Node(3);           root.left.left = new Node(4);      root.left.right = new Node(5);     root.right.left = new Node(6);root.right.right = new Node(7); createLoops(root); // traversing odd loop from any// random odd nodedocument.write("Odd nodes: ");traverseLoop(root.right); document.write( "<br>Even nodes: "); // traversing even loop from any// random even nodetraverseLoop(root.left);       // This code is contributed by patel2127 </script>
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 write.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. 
mohit kumar 29
aashish1995
GauravRajput1
patel2127
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