Binary Tree Iterator for Inorder Traversal
Given a Binary Tree and an input array. The task is to create an Iterator that utilizes next() and hasNext() functions to perform Inorder traversal on the binary tree.
Examples:
Input: 8 Input Array = [next(), hasNext(), next(), next(), next(), hasNext(), next(), next(), hasNext()]
/ \
3 9
/ \
2 4
\
5
Output: [2, true, 3, 4, 5, true, 8, 9, false]
Explanation: According to in order traversal answer to the input array is calculated.
Inorder traversal = {2, 3, 4, 5, 8, 9}
Input: 4 Input Array = [hasNext(), next(), next(), hasNext()]
/ \
3 2
\
1
Output: [true, 3, 4 true]
Naive Approach: A way is needed to traverse back to the ancestor once we reach the leaf node of the binary tree. A Stack data structure can be used for this.
Algorithm:
Class is Instantiated
- initialize the stack
- set current node = root
- while current != NULL
- add current to stack
- current = current.left
hasNext() function
IF the stack is not empty
return true
ELSE
return false
next() function
- IF stack is empty (or hasNext() returns false)
- ELSE
- Initialize current = stack.top
- Pop the element from the stack
- If current.right != NULL
- Initialize next = current->right
- while next != NULL
- add next to the stack
- next = next.left
- return current
Below is the implementation of above approach
C++
#include <iostream>
#include <stack>
using namespace std;
struct Node {
int data;
Node* left;
Node* right;
};
Node* newNode( int data)
{
Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return node;
}
class InorderIterator {
private :
stack<Node*> traversal;
public :
InorderIterator(Node* root)
{
moveLeft(root);
}
void moveLeft(Node* current)
{
while (current) {
traversal.push(current);
current = current->left;
}
}
bool hasNext()
{
return !traversal.empty();
}
Node* next()
{
if (!hasNext())
throw "No such element Exists" ;
Node* current = traversal.top();
traversal.pop();
if (current->right)
moveLeft(current->right);
return current;
}
};
int main()
{
Node* root = newNode(8);
root->right = newNode(9);
root->left = newNode(3);
root->left->left = newNode(2);
root->left->right = newNode(4);
root->left->right->right = newNode(5);
InorderIterator itr(root);
try {
cout << itr.next()->data << " " ;
cout << itr.hasNext() << " " ;
cout << itr.next()->data << " " ;
cout << itr.next()->data << " " ;
cout << itr.next()->data << " " ;
cout << itr.hasNext() << " " ;
cout << itr.next()->data << " " ;
cout << itr.next()->data << " " ;
cout << itr.hasNext() << " " ;
}
catch ( const char * msg) {
cout << msg;
}
return 0;
}
|
Java
import java.util.*;
class Node {
int data;
Node left;
Node right;
Node( int data)
{
this .data = data;
left = right = null ;
}
}
class InorderIterator {
private Stack<Node> traversal;
InorderIterator(Node root)
{
traversal = new Stack<Node>();
moveLeft(root);
}
private void moveLeft(Node current)
{
while (current != null ) {
traversal.push(current);
current = current.left;
}
}
public boolean hasNext()
{
return !traversal.isEmpty();
}
public Node next()
{
if (!hasNext())
throw new NoSuchElementException();
Node current = traversal.pop();
if (current.right != null )
moveLeft(current.right);
return current;
}
}
class Test {
public static void main(String args[])
{
Node root = new Node( 8 );
root.right = new Node( 9 );
root.left = new Node( 3 );
root.left.left = new Node( 2 );
root.left.right = new Node( 4 );
root.left.right.right = new Node( 5 );
InorderIterator itr = new InorderIterator(root);
try {
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
}
catch (NoSuchElementException e) {
System.out.print( "No such element Exists" );
}
}
}
|
Python
class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
class InorderIterator:
def __init__( self , root):
self .traversal = []
self .moveLeft(root)
def moveLeft( self , current):
while current ! = None :
self .traversal.append(current)
current = current.left
def hasNext( self ):
return len ( self .traversal) > 0
def next ( self ):
if not self .hasNext():
raise Exception( 'No such element Exists' )
current = self .traversal.pop()
if current.right ! = None :
self .moveLeft(current.right)
return current
root = Node( 8 )
root.right = Node( 9 )
root.left = Node( 3 )
root.left.left = Node( 2 )
root.left.right = Node( 4 )
root.left.right.right = Node( 5 )
itr = InorderIterator(root)
try :
print (itr. next ().data)
print (itr.hasNext())
print (itr. next ().data)
print (itr. next ().data)
print (itr. next ().data)
print (itr.hasNext())
print (itr. next ().data)
print (itr. next ().data)
print (itr.hasNext())
except Exception as e:
print ( "No such element Exists" )
|
C#
using System;
using System.Collections.Generic;
class Node
{
public int Data { get ; set ; }
public Node Left { get ; set ; }
public Node Right { get ; set ; }
public Node( int data)
{
Data = data;
Left = Right = null ;
}
}
class InorderIterator
{
private Stack<Node> traversal;
public InorderIterator(Node root)
{
traversal = new Stack<Node>();
MoveLeft(root);
}
private void MoveLeft(Node current)
{
while (current != null )
{
traversal.Push(current);
current = current.Left;
}
}
public bool HasNext()
{
return traversal.Count > 0;
}
public Node Next()
{
if (!HasNext())
throw new InvalidOperationException( "No more elements in the inorder traversal." );
Node current = traversal.Pop();
if (current.Right != null )
MoveLeft(current.Right);
return current;
}
}
class GFG
{
static void Main( string [] args)
{
Node root = new Node(8);
root.Right = new Node(9);
root.Left = new Node(3);
root.Left.Left = new Node(2);
root.Left.Right = new Node(4);
root.Left.Right.Right = new Node(5);
InorderIterator itr = new InorderIterator(root);
try
{
Console.Write(itr.Next().Data + " " );
Console.Write(itr.HasNext() + " " );
Console.Write(itr.Next().Data + " " );
Console.Write(itr.Next().Data + " " );
Console.Write(itr.Next().Data + " " );
Console.Write(itr.HasNext() + " " );
Console.Write(itr.Next().Data + " " );
Console.Write(itr.Next().Data + " " );
Console.Write(itr.HasNext() + " " );
}
catch (InvalidOperationException e)
{
Console.Write( "No such element Exists" );
}
}
}
|
Javascript
class Node {
constructor(data) {
this .data = data;
this .left = null ;
this .right = null ;
}
}
class InorderIterator {
constructor(root) {
this .traversal = [];
this .moveLeft(root);
}
moveLeft(current) {
while (current != null ) {
this .traversal.push(current);
current = current.left;
}
}
hasNext() {
return this .traversal.length > 0;
}
next() {
if (! this .hasNext())
throw new Error( 'No such element Exists' );
let current = this .traversal.pop();
if (current.right != null )
this .moveLeft(current.right);
return current;
}
}
let root = new Node(8);
root.right = new Node(9);
root.left = new Node(3);
root.left.left = new Node(2);
root.left.right = new Node(4);
root.left.right.right = new Node(5);
let itr = new InorderIterator(root);
try {
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
}
catch (e) {
console.log( "No such element Exists" );
}
|
Output
2 true 3 4 5 true 8 9 false
Time Complexity: O(N), Where N is the number of nodes in the binary tree.
Auxiliary Space: O(N), The Stack will hold all N elements in the worst case.
Efficient Approach: Morris Traversal can be used to solve this question using constant space. The idea behind morris traversal is to create a temporary link between a node and the right-most node in its left sub-tree so that the ancestor node can be backtracked. A reference of the ancestor node is set to the right child of the right-most node in its left sub-tree.
Algorithm:
Class is Instantiated
Initialize current = root and rightMost = NULL
hasNext() function
IF current != NULL
return true
ELSE
return false
next() function
- IF current = NULL ( or hasNext() returns false)
- ELSE
- IF current.left = NULL
- Initialize temp = current
- current = current.right
- return temp
- ELSE
- Initialize rightMost = current->left
- while rightMost.right != NULL && rightMost.right != current
- rightMost = rightMost.right
- IF rightMost.right == NULL
- rightMost.right = current
- current = current.left
- ELSE
- temp = current
- rightMost.right = null
- current = current.right
- return current
- Call the function again
Below is the implementation of above approach.
C++
#include <iostream>
#include <stack>
struct Node {
int data;
Node *left, *right;
Node( int data) {
this ->data = data;
this ->left = this ->right = NULL;
}
};
class InorderIterator {
private :
Node *current, *rightMost;
public :
InorderIterator(Node *root) {
current = root;
rightMost = NULL;
}
bool hasNext() { return current != NULL; }
Node* next() {
if (!hasNext()) {
std::cout << "No such element exists" << std::endl;
return NULL;
}
if (current->left == NULL) {
Node *temp = current;
current = current->right;
return temp;
}
rightMost = current->left;
while (rightMost->right != NULL
&& rightMost->right != current) {
rightMost = rightMost->right;
}
if (rightMost->right == NULL) {
rightMost->right = current;
current = current->left;
} else {
rightMost->right = NULL;
Node *temp = current;
current = current->right;
return temp;
}
return next();
}
};
int main() {
Node *root = new Node(8);
root->right = new Node(9);
root->left = new Node(3);
root->left->left = new Node(2);
root->left->right = new Node(4);
root->left->right->right = new Node(5);
InorderIterator itr(root);
std::cout << itr.next()->data << " " ;
std::cout << itr.hasNext() << " " ;
std::cout << itr.next()->data << " " ;
std::cout << itr.next()->data << " " ;
std::cout << itr.next()->data << " " ;
std::cout << itr.hasNext() << " " ;
std::cout << itr.next()->data << " " ;
std::cout << itr.next()->data << " " ;
std::cout << itr.hasNext() << " " ;
return 0;
}
|
Java
import java.util.*;
class Node {
int data;
Node left;
Node right;
Node( int data)
{
this .data = data;
left = right = null ;
}
}
class InorderIterator {
private Node current, rightMost;
InorderIterator(Node root)
{
current = root;
rightMost = null ;
}
public boolean hasNext() { return current != null ; }
public Node next()
{
if (!hasNext())
throw new NoSuchElementException();
if (current.left == null ) {
Node temp = current;
current = current.right;
return temp;
}
rightMost = current.left;
while (rightMost.right != null
&& rightMost.right != current)
rightMost = rightMost.right;
if (rightMost.right == null ) {
rightMost.right = current;
current = current.left;
}
else {
rightMost.right = null ;
Node temp = current;
current = current.right;
return temp;
}
return next();
}
}
class Test {
public static void main(String args[])
{
Node root = new Node( 8 );
root.right = new Node( 9 );
root.left = new Node( 3 );
root.left.left = new Node( 2 );
root.left.right = new Node( 4 );
root.left.right.right = new Node( 5 );
InorderIterator itr = new InorderIterator(root);
try {
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.next().data + " " );
System.out.print(itr.hasNext() + " " );
}
catch (NoSuchElementException e) {
System.out.print( "No such element Exists" );
}
}
}
|
Python
class Node:
def __init__( self , data):
self .data = data
self .left = self .right = None
class InorderIterator:
def __init__( self , root):
self .current = root
self .right_most = None
def has_next( self ):
return self .current is not None
def next ( self ):
if not self .has_next():
print ( "No such element exists" )
return None
if self .current.left is None :
temp = self .current
self .current = self .current.right
return temp
self .right_most = self .current.left
while self .right_most.right is not None and self .right_most.right ! = self .current:
self .right_most = self .right_most.right
if self .right_most.right is None :
self .right_most.right = self .current
self .current = self .current.left
else :
self .right_most.right = None
temp = self .current
self .current = self .current.right
return temp
return self . next ()
root = Node( 8 )
root.right = Node( 9 )
root.left = Node( 3 )
root.left.left = Node( 2 )
root.left.right = Node( 4 )
root.left.right.right = Node( 5 )
itr = InorderIterator(root)
print (itr. next ().data)
print (itr.has_next())
print (itr. next ().data)
print (itr. next ().data)
print (itr. next ().data)
print (itr.has_next())
print (itr. next ().data)
print (itr. next ().data)
print (itr.has_next())
|
Javascript
class Node {
constructor(data) {
this .data = data;
this .left = null ;
this .right = null ;
}
}
class InorderIterator {
constructor(root) {
this .current = root;
this .rightMost = null ;
}
hasNext() { return this .current != null ; }
next() {
if (! this .hasNext())
throw new Error( "No such element Exists" );
if ( this .current.left == null ) {
let temp = this .current;
this .current = this .current.right;
return temp;
}
this .rightMost = this .current.left;
while ( this .rightMost.right != null
&& this .rightMost.right != this .current)
this .rightMost = this .rightMost.right;
if ( this .rightMost.right == null ) {
this .rightMost.right = this .current;
this .current = this .current.left;
}
else {
this .rightMost.right = null ;
let temp = this .current;
this .current = this .current.right;
return temp;
}
return this .next();
}
}
let root = new Node(8);
root.right = new Node(9);
root.left = new Node(3);
root.left.left = new Node(2);
root.left.right = new Node(4);
root.left.right.right = new Node(5);
let itr = new InorderIterator(root);
try {
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
console.log(itr.next().data + " " );
console.log(itr.next().data + " " );
console.log(itr.hasNext() + " " );
}
catch (e) {
console.log( "No such element Exists" );
}
|
C#
using System;
using System.Collections.Generic;
public class Node {
public int data;
public Node left, right;
public Node( int data)
{
this .data = data;
left = right = null ;
}
}
public class InorderIterator {
private Node current, rightMost;
public InorderIterator(Node root)
{
current = root;
rightMost = null ;
}
public bool HasNext() { return current != null ; }
public Node Next()
{
if (!HasNext())
throw new Exception( "No such element exists" );
if (current.left == null ) {
Node temp = current;
current = current.right;
return temp;
}
rightMost = current.left;
while (rightMost.right != null
&& rightMost.right != current)
rightMost = rightMost.right;
if (rightMost.right == null ) {
rightMost.right = current;
current = current.left;
}
else {
rightMost.right = null ;
Node temp = current;
current = current.right;
return temp;
}
return Next();
}
}
public class GFG {
static public void Main()
{
Node root = new Node(8);
root.right = new Node(9);
root.left = new Node(3);
root.left.left = new Node(2);
root.left.right = new Node(4);
root.left.right.right = new Node(5);
InorderIterator itr = new InorderIterator(root);
try {
Console.Write(itr.Next().data + " " );
Console.Write(itr.HasNext() + " " );
Console.Write(itr.Next().data + " " );
Console.Write(itr.Next().data + " " );
Console.Write(itr.Next().data + " " );
Console.Write(itr.HasNext() + " " );
Console.Write(itr.Next().data + " " );
Console.Write(itr.Next().data + " " );
Console.Write(itr.HasNext() + " " );
}
catch (Exception e) {
Console.Write( "No such element Exists" );
}
}
}
|
Output
2 true 3 4 5 true 8 9 false
Time Complexity: O(N), where N is the number of nodes in the binary tree. Although we are creating temporary links are created and nodes are traversed multiple times (at most 3 times), the time complexity is still linear.
Auxiliary Space: O(1)
Last Updated :
20 Feb, 2023
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