BST Iterator 2
You are given a class named as BSTIterator that represents an iterator over inorder traversal of a binary search tree. You need to implement the following things as follows:

1. BSTIterator(Node root) - It is a parameterized constructor in which you are given the root of the Binary search tree. It will be called whenever an object of BSTIterator is created. 2. next() - This member function will return the next element in the in-order traversal of the binary search tree. You need to implement this function. 3. hasNext() - This function will return true if there exists the next element in the traversal else it will return false. You need to implement this function 4. prev() - This member function will return the previous element in the in-order traversal of the binary search tree. You need to implement this function. 5. hasPrev() - This function will return true if there exists the previous element in the traversal else it will return false. You need to implement this function

The binary search tree has ‘N’ nodes and you need to print the inorder traversal of the tree using the iterator.

Input Format:

The first line of the input contains an integer ‘T’ denoting the number of test cases. The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. For example, the input for the tree depicted in the below image would be :

4 2 6 1 3 5 7 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 4 Level 2 : Left child of 4 = 2 Right child of 4 = 6 Level 3 : Left child of 2 = 1 Right child of 2 = 3 Left child of 6 = 5 Right child of 6 = 7 Level 4 : Left child of 1 = null (-1) Right child of 1 = null (-1) Left child of 3 = null (-1) Right child of 3 = null (-1) Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 4 2 6 1 3 5 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case print a single line containing space-separated integers denoting the inorder traversal of the binary search tree. The output of each test case will be printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1

Question

BST Iterator 2

You are given a class named as BSTIterator that represents an iterator over inorder traversal of a binary search tree. You need to implement the following things as follows:

1. BSTIterator(Node root) - It is a parameterized constructor in which you are given the root of the Binary search tree. It will be called whenever an object of BSTIterator is created.

2. next() - This member function will return the next element in the in-order traversal of the binary search tree. You need to implement this function.

3. hasNext() - This function will return true if there exists the next element in the traversal else it will return false. You need to implement this function

4. prev() - This member function will return the previous element in the in-order traversal of the binary search tree. You need to implement this function.

5. hasPrev() - This function will return true if there exists the previous element in the traversal else it will return false. You need to implement this function

The binary search tree has ‘N’ nodes and you need to print the inorder traversal of the tree using the iterator.

Input Format:

The first line of the input contains an integer ‘T’ denoting the number of test cases.

The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

4
2 6
1 3 5 7
-1 -1 -1 -1 -1 -1

Explanation :

Level 1 :
The root node of the tree is 4

Level 2 :
Left child of 4 = 2
Right child of 4 = 6

Level 3 :
Left child of 2 = 1
Right child of 2 = 3
Left child of 6 = 5
Right child of 6 = 7

Level 4 :
Left child of 1 = null (-1)
Right child of 1 = null (-1)
Left child of 3 = null (-1)
Right child of 3 = null (-1)
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree.

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

4 2 6 1 3 5 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case print a single line containing space-separated integers denoting the inorder traversal of the binary search tree.

The output of each test case will be printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 10
1 <= N <= 10 ^ 4  
1 <= A[i] <= 10 ^ 9 

Where ‘T’ is the number of test cases, ‘N’ is the number of nodes, and A[i] is the value of a node.

Time Limit: 1 sec.

Accepted Answer

Doing preprocessing

We need to implement four functions next(), hasNext(), prev() and hasPrev(). The next() function will move the iterator to the right and return the next element, the hasNext() function will return true if there exists the next element else false, the prev() function will move the iterator to the left and return the previous element, the hasPrev() function will return true if there exists the previous element else false.
We can prepare a vector that stores all the elements of the binary search tree. We can do so by performing inorder traversal on the given binary search tree and store the elements in the list.
We maintain a variable to iterate over the list that we prepared in the previous step.
- Now for each call of the next() function, we simply move on the list one step towards the right and return the value stored therein the array.
- For each call of the prev() function, we simply move on the list one step towards the left and return the value stored therein the array.
- For the call of the hasNext() function, we need to check if our iterator reached the end of the array or not. If it is reached to the end of the array that means no element is left we iterate over all the elements so we return false else we will return true.
- Similarly, for the call of the hasPrev() function, we need to check if our iterator is at the start of the array or not. If it is at the start of the array that means no element is present before that element so we return false else we will return true.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the binary search tree.

We are using a deque of size ‘N’ to store elements of the binary search tree.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the binary search tree.

We are traversing each node one time while performing inorder traversal that will take O(N) time complexity.

Each call of the next() function will work in O(1) complexity.

Each call of the hasNext() function will work in O(1) complexity.

Each call of the prev() function will work in O(1) complexity.

Each call of the hasPrev() function will work in O(1) complexity.

Thus, the final time complexity is O(N).

Python (3.5)

'''

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.

'''



class   TreeNode :

    def __init__(self, val) :

        self.val = val

        self.left = None

        self.right = None



    def __del__(self):

        if self.left:

            del self.left

        if self.right:

            del self.right

            

class BSTiterator:

    def __init__(self,root):

        self.index = 0

        self.nodeValues = []

        self.inorder(root)



    def next(self):

        next = self.nodeValues[self.index]

        

        #  Increment the index for next smallest element.

        self.index += 1

        return next



    def hasNext(self):

        

        if self.index < len(self.nodeValues):

            return True

        return False



    def prev(self):

        prev = self.nodeValues[self.index]

        

        # Decrement the index for prev element.

        self.index -= 1

        return prev



    def hasPrev(self):

        

        if self.index > len(self.nodeValues):

            return True

        return False

    

    # This function will used to perform inorder traversal on the BST.

    def inorder(self, root):

        if not root:

            return 

        

        self.inorder(root.left)

        self.nodeValues.append(root.data)

        self.inorder(root.right)

C++ (g++ 5.4)

/*

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.



*/



class BSTiterator

{

    public:

    

    // Create a vector to store node's values.

    vector nodeValues;

    

    // This will be used to iterate over the list.

    int index = 0;

    

    BSTiterator(TreeNode *root)

    {

        index = 0;

        inorder(root);

    }



    int next()

    {

        int next = nodeValues[index];

        

        // Increment the index for next smallest element.

        ++index;

        return next;

    }



    bool hasNext()

    {

        

        if (index < nodeValues.size())

        {

            return true;

        }

        else

        {

            return false;

        }

    }



	int prev()

    {

        int prev = nodeValues[index];

        

        // Decrement the index for prev element.

        --index;

        return prev;

    }



    bool hasPrev()

    {

        

        if (index > nodeValues.size())

        {

            return true;

        }

        else

        {

            return false;

        }

    }

    

    // This function will used to perform inorder traversal on the BST.

    void inorder(TreeNode *root)

    {

        if (root == NULL)

        {

            return;

        }

        

        // Traverse on the left child.

        inorder(root->left);

        

        // Store the data.

        nodeValues.push_back(root->data);

        

        // Traverse on the right child.

        inorder(root->right);

    }

};

Java (SE 1.8)

/*

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.



*/



import java.util.ArrayList;



public class BSTiterator {



	// Create a ArrayList to store node's values.

	private ArrayList nodeValues;



	// This will be used to iterate over the list.

	private int index = 0;



	public BSTiterator(TreeNode root) {

		index = 0;

		nodeValues = new ArrayList();

		inorder(root);

	}



	public int next() {

		int next = nodeValues.get(index);



		// Increment the index for next smallest element.

		++index;

		return next;

	}



	public boolean hasNext() {



		if (index < nodeValues.size()) {

			return true;

		} else {

			return false;

		}

	}



	public int prev() {

		int prev = nodeValues.get(index);



		// Decrement the index for prev element.

		--index;

		return prev;

	}



	public boolean hasPrev() {



		if (index > nodeValues.size()) {

			return true;

		} else {

			return false;

		}

	}



	// This function will used to perform inorder traversal on the BST.

	private void inorder(TreeNode root) {

		if (root == null) {

			return;

		}



		// Traverse on the left child.

		inorder(root.left);



		// Store the data.

		nodeValues.add(root.data);



		// Traverse on the right child.

		inorder(root.right);

	}

};

Accepted Answer

Using Deque

Instead of storing all the elements in the list. We can prepare a deque in which we store the next element on its front and previous element on its back. We update the deque on demand to maintain the next and previous element on its front and back.
We know that for the given root of the binary search tree its smallest element will be the leftmost node’s value in its subtree. So first of all we insert all nodes in the deque till we reach the leftmost node in the tree, which means that our stack now will store the smallest element on its front.
When we call the next() function for the first time we need to return the element which is already present at our deque’s front. We will pop the front value from the deque and return it. But before returning the value we need to ensure that our deque will contain the next element on its front for further calls of the next() function. So there are two cases we need to check to maintain our deque :
- First, check if the node we are popping currently is a leaf node or not. If it is a leaf node then in that case we simply pop it because the next top contains the next element already.
- If the node we are popping is not a leaf then we need to check if it has the right child or not. We don’t need to check the left child because the way we are maintaining the deque, the left child of the node was already processed. So for the right child, we again add all leftmost nodes present in the subtree of the right child including itself in the deque.
Similarly, when we call the prev() function for the first time we need to return the element which is already present at our deque’s back. We will pop the back value from the deque and return it. But before returning the value we need to ensure that our deque will contain the previous element on its back for further calls of the prev() function. So there are two cases we need to check to maintain our deque :
- First, check if the node we are popping currently is a leaf node or not. If it is a leaf node then in that case we simply pop it because the next back contains the previous element already.
- If the node we are popping is not a leaf then we need to check if it has the left child or not. We don’t need to check the right child because the way we are maintaining the deque, the left child of the node was already processed. So for the left child, we again add all rightmost nodes present in the subtree of the left child including itself in the deque.
If you observe we are traversing the binary search tree according to the inorder traversal and maintaining the deque to store the nodes.
For the call of hasNext() function we just need to check if the deque is empty or not. If it is empty that means we processed all the nodes so we return false else we will return true.
For the call of hasPrev() function we just need to check if the deque is empty or not. If it is empty that means we processed all the nodes so we return false else we will return true.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the binary search tree.

We are using a deque of size ‘N’ to store elements of the binary search tree.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the binary search tree.

We are traversing each node one time while performing inorder traversal that will take O(N) time complexity.

Each call of the next() function will work in O(1) complexity.

Each call of the hasNext() function will work in O(1) complexity.

Each call of the prev() function will work in O(1) complexity.

Each call of the hasPrev() function will work in O(1) complexity.

Thus, the final time complexity is O(N).

Java (SE 1.8)

/*

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.

*/



import java.util.Deque;

import java.util.LinkedList;



public class BSTiterator {



	// Create a deque which will store the elements.

	private Deque> deq;



	BSTiterator(TreeNode root) {



		// Fill the deque with leftmost nodes present in the subtree of root.

		deq = new LinkedList>();

		leftMostInorder(root);

	}



	public int next() {



		// Pop the element.

		TreeNode front = deq.getFirst();

		deq.removeFirst();



		// Check if it has right child.

		if (front.right != null) {



			// Push leftmost nodes present in the subtree of right child.

			leftMostInorder(front.right);

		}



		return front.data;

	}



	public boolean hasNext() {



		// If size of deque is greater than zero that means there are still some nodes left for processing.

		if (deq.size() > 0) {

			return true;

		} else {

			return false;

		}

	}



	public int prev() {



		// Pop the element.

		TreeNode back = deq.getLast();

		deq.removeLast();



		// Check if it has left child.

		if (back.left != null) {



			// Push leftmost nodes present in the subtree of left child.

			leftMostInorder(back.left);

		}

		return back.data;

	}



	public boolean hasPrev() {

		// If size of deque is less than zero that means there are still some nodes left for processing.

		if (deq.size() < 0) {

			return true;

		} else {

			return false;

		}

	}



	private void leftMostInorder(TreeNode root) {

		while (root != null) {

			deq.addFirst(root);

			root = root.left;

		}

	}

};

C++ (g++ 5.4)

/*

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.

*/



#include 



class BSTiterator

{

public:



    // Create a deque which will store the elements.

    deque *> deq;



    BSTiterator(TreeNode *root)

    {



        // Fill the deque with leftmost nodes present in the subtree of root.

        leftMostInorder(root);

    }



    int next()

    {



        // Pop the element.

        TreeNode *front = deq.front();

        deq.pop_front();



        // Check if it has right child.

        if (front->right != NULL)

        {



            // Push leftmost nodes present in the subtree of right child.

            leftMostInorder(front->right);

        }



        return front->data;

    }



    bool hasNext()

    {



        // If size of deque is greater than zero that means there are still some nodes left for processing.

        if (deq.size() > 0)

        {

            return true;

        }

        else

        {

            return false;

        }

    }

    

    int prev()

    {



        // Pop the element.

        TreeNode *back = deq.back();

        deq.pop_back();



        // Check if it has left child.

        if (back->left != NULL)

        {



            // Push leftmost nodes present in the subtree of left child.

            leftMostInorder(back->left);

        }

        return back->data;

    }



    bool hasPrev()

    {

        // If size of deque is less than zero that means there are still some nodes left for processing.

        if (deq.size() < 0)

        {

            return true;

        }

        else

        {

            return false;

        }

    }



    // This function will used to push leftmost nodes in the deque present in the subtree of root.

    void leftMostInorder(TreeNode *root)

    {

        while (root != NULL)

        {

            deq.push_front(root);

            root = root->left;

        }

    }

};

Python (3.5)

'''

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.

'''



class   TreeNode :

    def __init__(self, val) :

        self.val = val

        self.left = None

        self.right = None



    def __del__(self):

        if self.left:

            del self.left

        if self.right:

            del self.right

            

class BSTiterator:

    def __init__(self,root):

        self.deq = []

        

        # Fill the deque with leftmost nodes present in the subtree of root.

        self.leftMostInorder(root)



    def next(self):

        # Pop the element.

        front = self.deq.pop(0)

        

        if front.right:

            self.leftMostInorder(front.right)

            

        return front.data



    def hasNext(self):

        

        # If size of deque is greater than zero that means there are still some nodes left for processing.

        if len(self.deq) > 0:

            return True

        return False



    def prev(self):

        # Pop the element.

        back = self.deq.pop()

        

        if back.left:

            self.leftMostInorder(back.left)

            

        return back.data



    def hasPrev(self):

        # If size of deque is less than zero that means there are still some nodes left for processing.

        if len(self.deq) < 0:

            return True

        return False

    

    # This function will used to push leftmost nodes in the deque present in the subtree of root.

    def leftMostInorder(self, root):

        while root:

            self.deq = [root] + self.deq

            root = root.left

You are given a class named as BSTIterator that represents an iterator over inorder traversal of a binary search tree. You need to implement the following things as follows:

The binary search tree has ‘N’ nodes and you need to print the inorder traversal of the tree using the iterator.

Input Format:

Explanation :

Note :

Output Format:

Note:

Constraints:

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