Given a binary tree. Make a modification so that after the modification, it can be traversed in order using only the correct pointers. During modification, you can use both the right and left pointers.
Example:
Input : 10
/ \
8 2
/ \
3 5
Output : 10
\
8
\
3
\
5
\
2
Explanation : The preorder traversal
of given binary tree is 10 8 3 5 2.Recommended: Try the following {IDE} first, before proceeding with the solution.
Method 1 (Recursive)
You need to make the right pointer of the root point to the left subtree.
If the node happens to have a left child node, you can simply move the child node to the right to complete the processing of the node.
If there are also suitable children, then they should
The rightmost right child of the original left subtree.
The above function used in the code processes a node and then returns the rightmost node of the transformed subtree.
C ++
//C code to modify binary tree for
//traversal using only right pointer
#include <bits/stdc++.h>
using namespace std;
//A binary tree node has data, //left child and right child
struct Node {
int data;
struct Node* left;
struct Node* right;
};
//function that allocates a new node
//with the given data and NULL left
//and right pointers.
struct Node* newNode( int data)
{
struct Node* node = new struct Node;
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
//Function to modify tree
struct Node* modifytree( struct Node* root)
{
struct Node* right = root->right;
struct Node* rightMost = root;
//if the left tree exists
if (root->left) {
//get the right-most of the
//original left subtree
rightMost = modifytree(root->left);
//set root right to left subtree
root->right = root->left;
root->left = NULL;
}
//if the right subtree does
//not exists we are done!
if (!right)
return rightMost;
//set right pointer of right-most
//of the original left subtree
rightMost->right = right;
//modify the rightsubtree
rightMost = modifytree(right);
return rightMost;
}
//printing using right pointer only
void printpre( struct Node* root)
{
while (root != NULL) {
cout <<root->data <<" " ;
root = root->right;
}
}
//Driver program to test above functions
int main()
{
/* Constructed binary tree is
10
/\
8 2
/\
3 5 */
struct Node* root = newNode(10);
root->left = newNode(8);
root->right = newNode(2);
root->left->left = newNode(3);
root->left->right = newNode(5);
modifytree(root);
printpre(root);
return 0;
}Java
//Java code to modify binary tree for
//traversal using only right pointer
class GFG
{
//A binary tree node has data, //left child and right child
static class Node
{
int data;
Node left;
Node right;
};
//function that allocates a new node
//with the given data and null left
//and right pointers.
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = null ;
node.right = null ;
return (node);
}
//Function to modify tree
static Node modifytree( Node root)
{
Node right = root.right;
Node rightMost = root;
//if the left tree exists
if (root.left != null )
{
//get the right-most of the
//original left subtree
rightMost = modifytree(root.left);
//set root right to left subtree
root.right = root.left;
root.left = null ;
}
//if the right subtree does
//not exists we are done!
if (right == null )
return rightMost;
//set right pointer of right-most
//of the original left subtree
rightMost.right = right;
//modify the rightsubtree
rightMost = modifytree(right);
return rightMost;
}
//printing using right pointer only
static void printpre( Node root)
{
while (root != null )
{
System.out.print( root.data + " " );
root = root.right;
}
}
//Driver cde
public static void main(String args[])
{
/* Constructed binary tree is
10
/\
8 2
/\
3 5 */
Node root = newNode( 10 );
root.left = newNode( 8 );
root.right = newNode( 2 );
root.left.left = newNode( 3 );
root.left.right = newNode( 5 );
modifytree(root);
printpre(root);
}
}
//This code is contributed by Arnab KunduPython3
# Python code to modify binary tree for
# traversal using only right pointer
class newNode():
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# Function to modify tree
def modifytree(root):
right = root.right
rightMost = root
# if the left tree exists
if (root.left):
# get the right-most of the
# original left subtree
rightMost = modifytree(root.left)
# set root right to left subtree
root.right = root.left
root.left = None
# if the right subtree does
# not exists we are done!
if ( not right):
return rightMost
# set right pointer of right-most
# of the original left subtree
rightMost.right = right
# modify the rightsubtree
rightMost = modifytree(right)
return rightMost
# printing using right pointer only
def printpre(root):
while (root ! = None ):
print (root.data, end = " " )
root = root.right
# Driver code
if __name__ = = '__main__' :
""" Constructed binary tree is
10
/\
8 2
/\
3 5 """
root = newNode( 10 )
root.left = newNode( 8 )
root.right = newNode( 2 )
root.left.left = newNode( 3 )
root.left.right = newNode( 5 )
modifytree(root)
printpre(root)
# This code is contributed by SHUBHAMSINGH10C#
//C# code to modify binary tree for
//traversal using only right pointer
using System;
class GFG
{
//A binary tree node has data, //left child and right child
public class Node
{
public int data;
public Node left;
public Node right;
};
//function that allocates a new node
//with the given data and null left
//and right pointers.
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = null ;
node.right = null ;
return (node);
}
//Function to modify tree
static Node modifytree( Node root)
{
Node right = root.right;
Node rightMost = root;
//if the left tree exists
if (root.left != null )
{
//get the right-most of the
//original left subtree
rightMost = modifytree(root.left);
//set root right to left subtree
root.right = root.left;
root.left = null ;
}
//if the right subtree does
//not exists we are done!
if (right == null )
return rightMost;
//set right pointer of right-most
//of the original left subtree
rightMost.right = right;
//modify the rightsubtree
rightMost = modifytree(right);
return rightMost;
}
//printing using right pointer only
static void printpre( Node root)
{
while (root != null )
{
Console.Write( root.data + " " );
root = root.right;
}
}
//Driver cde
public static void Main(String []args)
{
/* Constructed binary tree is
10
/\
8 2
/\
3 5 */
Node root = newNode(10);
root.left = newNode(8);
root.right = newNode(2);
root.left.left = newNode(3);
root.left.right = newNode(5);
modifytree(root);
printpre(root);
}
}
//This code is contributed by 29AjayKumarThe output is as follows:
10 8 3 5 2Method 2 (Iterative)
This can be easily done using iteration’s pre-order traversal. Look here.
Iterative traversal
The idea is to maintain a variable prev, which maintains the previous node of the pre-order traversal. Each time a new node is encountered, the node sets its permissions to the previous node, and the previous node equals the current node. Finally, we will get a kind of linked list whose first element is root, then left child, then right, and so on.
C ++
//C++ code to modify binary tree for
//traversal using only right pointer
#include <iostream>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
//A binary tree node has data, //left child and right child
struct Node {
int data;
struct Node* left;
struct Node* right;
};
//Helper function that allocates a new
//node with the given data and NULL
//left and right pointers.
struct Node* newNode( int data)
{
struct Node* node = new struct Node;
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
//An iterative process to set the right
//pointer of Binary tree
void modifytree( struct Node* root)
{
//Base Case
if (root == NULL)
return ;
//Create an empty stack and push root to it
stack<Node*> nodeStack;
nodeStack.push(root);
/* Pop all items one by one.
Do following for every popped item
a) print it
b) push its right child
c) push its left child
Note that right child is pushed first
so that left is processed first */
struct Node* pre = NULL;
while (nodeStack.empty() == false ) {
//Pop the top item from stack
struct Node* node = nodeStack.top();
nodeStack.pop();
//Push right and left children of
//the popped node to stack
if (node->right)
nodeStack.push(node->right);
if (node->left)
nodeStack.push(node->left);
//check if some previous node exists
if (pre != NULL) {
//set the right pointer of
//previous node to currrent
pre->right = node;
}
//set previous node as current node
pre = node;
}
}
//printing using right pointer only
void printpre( struct Node* root)
{
while (root != NULL) {
cout <<root->data <<" " ;
root = root->right;
}
}
//Driver code
int main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \
3 5
*/
struct Node* root = newNode(10);
root->left = newNode(8);
root->right = newNode(2);
root->left->left = newNode(3);
root->left->right = newNode(5);
modifytree(root);
printpre(root);
return 0;
}Java
//Java code to modify binary tree for
//traversal using only right pointer
import java.util.*;
class GfG {
//A binary tree node has data, //left child and right child
static class Node {
int data;
Node left;
Node right;
}
//Helper function that allocates a new
//node with the given data and NULL
//left and right pointers.
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = null ;
node.right = null ;
return (node);
}
//An iterative process to set the right
//pointer of Binary tree
static void modifytree(Node root)
{
//Base Case
if (root == null )
return ;
//Create an empty stack and push root to it
Stack<Node> nodeStack = new Stack<Node> ();
nodeStack.push(root);
/* Pop all items one by one.
Do following for every popped item
a) print it
b) push its right child
c) push its left child
Note that right child is pushed first
so that left is processed first */
Node pre = null ;
while (nodeStack.isEmpty() == false ) {
//Pop the top item from stack
Node node = nodeStack.peek();
nodeStack.pop();
//Push right and left children of
//the popped node to stack
if (node.right != null )
nodeStack.push(node.right);
if (node.left != null )
nodeStack.push(node.left);
//check if some previous node exists
if (pre != null ) {
//set the right pointer of
//previous node to currrent
pre.right = node;
}
//set previous node as current node
pre = node;
}
}
//printing using right pointer only
static void printpre(Node root)
{
while (root != null ) {
System.out.print(root.data + " " );
root = root.right;
}
}
//Driver code
public static void main(String[] args)
{
/* Constructed binary tree is
10
/\
8 2
/\
3 5
*/
Node root = newNode( 10 );
root.left = newNode( 8 );
root.right = newNode( 2 );
root.left.left = newNode( 3 );
root.left.right = newNode( 5 );
modifytree(root);
printpre(root);
}
}python
# Python code to modify binary tree for
# traversal using only right pointer
# A binary tree node has data, # left child and right child
class newNode():
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# An iterative process to set the right
# pointer of Binary tree
def modifytree( root):
# Base Case
if (root = = None ):
return
# Create an empty stack and append root to it
nodeStack = []
nodeStack.append(root)
''' Pop all items one by one.
Do following for every popped item
a) prit
b) append its right child
c) append its left child
Note that right child is appended first
so that left is processed first '''
pre = None
while ( len (nodeStack)):
# Pop the top item from stack
node = nodeStack[ - 1 ]
nodeStack.pop()
# append right and left children of
# the popped node to stack
if (node.right):
nodeStack.append(node.right)
if (node.left):
nodeStack.append(node.left)
# check if some previous node exists
if (pre ! = None ):
# set the right pointer of
# previous node to currrent
pre.right = node
# set previous node as current node
pre = node
# printing using right pointer only
def printpre( root):
while (root ! = None ):
print (root.data, end = " " )
root = root.right
# Driver code
''' Constructed binary tree is
10
/\
8 2
/\
3 5
'''
root = newNode( 10 )
root.left = newNode( 8 )
root.right = newNode( 2 )
root.left.left = newNode( 3 )
root.left.right = newNode( 5 )
modifytree(root)
printpre(root)
# This code is contributed by SHUBHAMSINGH10C#
//C# code to modify binary tree for
//traversal using only right pointer
using System;
using System.Collections;
class GfG
{
//A binary tree node has data, //left child and right child
public class Node
{
public int data;
public Node left;
public Node right;
}
//Helper function that allocates a new
//node with the given data and NULL
//left and right pointers.
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = null ;
node.right = null ;
return (node);
}
//An iterative process to set the right
//pointer of Binary tree
static void modifytree(Node root)
{
//Base Case
if (root == null )
return ;
//Create an empty stack and Push root to it
Stack nodeStack = new Stack();
nodeStack.Push(root);
/* Pop all items one by one.
Do following for every Popped item
a) print it
b) Push its right child
c) Push its left child
Note that right child is Pushed first
so that left is processed first */
Node pre = null ;
while (nodeStack.Count !=0)
{
//Pop the top item from stack
Node node = (Node)nodeStack.Peek();
nodeStack.Pop();
//Push right and left children of
//the Popped node to stack
if (node.right != null )
nodeStack.Push(node.right);
if (node.left != null )
nodeStack.Push(node.left);
//check if some previous node exists
if (pre != null )
{
//set the right pointer of
//previous node to currrent
pre.right = node;
}
//set previous node as current node
pre = node;
}
}
//printing using right pointer only
static void printpre(Node root)
{
while (root != null )
{
Console.Write(root.data + " " );
root = root.right;
}
}
//Driver code
public static void Main(String []args)
{
/* Constructed binary tree is
10
/\
8 2
/\
3 5
*/
Node root = newNode(10);
root.left = newNode(8);
root.right = newNode(2);
root.left.left = newNode(3);
root.left.right = newNode(5);
modifytree(root);
printpre(root);
}
}
//This code is contributed by
//Arnab KunduThe output is as follows:
10 8 3 5 2