BST iterator
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 smallest 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 smallest element in the traversal else it will return false. You need to implement this function

The binary search tree has ‘N’ nodes 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 and the only 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:

The single line of output for each test case should contain the inorder traversal of the binary search tree.

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

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 smallest 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 smallest element in the traversal else it will return false. You need to implement this function

The binary search tree has ‘N’ nodes 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 and the only 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:

The single line of output for each test case should contain the inorder traversal of the binary search tree.

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

I solved it by using stack and then performing an in order traversal of the BST.

Accepted Answer

Tree flattening

We need to implement two functions next() and hasNext(). The next() function will move the iterator to the right and return the next smaller element and the hasNext() function will return true if there exists the next smallest element else false.
We can prepare a list that stores all the elements of the binary search tree in sorted order. We can do so by performing inorder traversal on the given binary search tree and store the elements in the list. As the inorder traversal gives us a sorted order of elements so the list will contain the elements in sorted order.
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. As the array is sorted so it will be the next smaller element.
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.

Space Complexity: O(n)Explanation:

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

We are using a list 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.

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.List;

import java.util.ArrayList;



public class Solution {



    static class BSTIterator {



        // Create a list to store node's value in sorted order

        List nodeValues;

        

        // This will be used to iterate over the list

        int index;



        public BSTIterator(TreeNode root) {

            nodeValues = new ArrayList<>();

            index = 0;

        

            // Fill the list with inorder traversal

            inorder(root);

        }



        public int next() {

            int nextSmallest = nodeValues.get(index);

        

            // Increment the index for next smallest element

            ++index;

            return nextSmallest;

        }



        public boolean hasNext() {

        

            // If index is less than size of the list that means there are still some nodes left for processing

            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);

        }



    }

}

C++ (g++ 5.4)

/*

	Time Complexity: O(N)

	Space complexity: O(N)



	Where 'N' is the number of nodes.



*/



class BSTiterator

{

public:

	

	// Create a list to store node's value in sorted order

	vector nodeValues;

	

	// This will be used to iterate over the list

	int index;



	BSTiterator(TreeNode *root)

	{

		index = 0;

		inorder(root);

	}



	int next()

	{

		int nextSmallest = nodeValues[index];

	

		// Increment the index for next smallest element

		++index;

		return nextSmallest;

	}



	bool hasNext()

	{

	

		// If index is less than size of the list that means there are still some nodes left for processing

		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);

	}

};

Python (3.5)

'''

    Time Complexity: O(N)

    Space complexity: O(N)



    Where 'N' is the number of nodes.



'''



class BSTiterator:

    

    def __init__(self, root):

        

        # Create a list to store node's value in sorted order

        self.nodeValues = []

        

        # This will be used to iterate over the list

        self.index = 0;

        self.inorder(root)



    def next(self):

        nextSmallest = self.nodeValues[self.index]

        

        # Increment the index for next smallest element

        self.index += 1

        return nextSmallest



    def hasNext(self):

        

        # If index is less than size of the list that means there are still some nodes left for processing

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

            return True

        else:

            return False



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

    def inorder(self, root):

        if (root == None):

            return

            

        # Traverse on the left child

        self.inorder(root.left)

        

        # Store the data

        self.nodeValues.append(root.data)

        

        # Traverse on the right child

        self.inorder(root.right)

Accepted Answer

Stack

Instead of storing all the elements in the list. We can prepare a stack in which we store the next smallest element on its top. We update the stack on demand to maintain the next smallest element on its top.
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 stack till we reach the leftmost node in the tree, which means that our stack now will store the smallest element on its top.
When we call the next() function for the first time we need to return the smallest element which is already present at our stack’s top. We will pop the value from the stack and return it. But before returning the value we need to ensure that our stack will contain the next smaller element on its top for further calls of the next() function. So there are two cases we need to check to maintain our stack :
- 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 smaller 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 stack, 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 stack.
If you observe we are traversing the binary search tree according to the inorder traversal and maintaining the stack to store the nodes.
For the call of hasNext() function we just need to check if the stack is empty or not. If it is empty that means we processed all the nodes so we return false else we will return true.
The following image summarizes the above approach:

Space Complexity: O(n)Explanation:

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

We are using a stack that can store ‘N’ nodes in the worst case. Thus, the final space complexity is O(N).

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.

All the calls of the next() function will overall have O(N) complexity.

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

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

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 stack which will store smallest element at the top

    stack *> st;



    BSTiterator(TreeNode *root)

    {

    	

        // Fill the stack with leftmost nodes present in the subtree of root

        leftMostInorder(root);

    }



    int next()

    {

    	

        // Pop the minimum

        TreeNode *top = st.top();

        st.pop();

        

        // Check if it has right child

        if (top->right != NULL)

        {

        

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

            leftMostInorder(top->right);

        }

        return top->data;

    }



    bool hasNext()

    {

        

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

        if (st.size() > 0)

        {

            return true;

        }

        else

        {

            return false;

        }

    }



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

    void leftMostInorder(TreeNode *root)

    {

        while (root != NULL)

        {

            st.push(root);

            root = root->left;

        }

    }

};

Python (3.5)

'''

    Time Complexity: O(N)

    Space complexity: O(N)



    Where 'N' is the number of nodes.



'''



class BSTiterator:

    

    # Create a stack which will store smallest element at the top

    st = []



    def __init__(self, root):

        

        # Fill the stack with leftmost nodes present in the subtree of root

        self.leftMostInorder(root)

    



    def next(self):



        # Pop the minimum

        top = self.st[-1]

        self.st.pop()

        

        # Check if it has right child

        if (top.right != None):

        

            # Push leftmost nodes present in the subtree of right child

            self.leftMostInorder(top.right)

        

        return top.data

    

    def hasNext(self):

    

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

        if (len(self.st) > 0):

            return True

        else:

            return False



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

    def leftMostInorder(self, root):

        while (root != None):

            self.st.append(root)

            root = root.left;

Java (SE 1.8)

/*

    Time Complexity: O(N)

    Space complexity: O(N)



    Where 'N' is the number of nodes.



 */



import java.util.Stack;



public class Solution {



    static class BSTIterator {



        // Create a stack which will store smallest element at the top

        Stack> st;



        public BSTIterator(TreeNode root) {

            st = new Stack<>();

            

            // Fill the stack with leftmost nodes present in the subtree of root

            leftMostInorder(root);

        }



        public int next() {

            

            // Pop the minimum

            TreeNode top = st.pop();

            

            // Check if it has right child

            if (top.right != null) {

            

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

                leftMostInorder(top.right);

            }

            return top.data;

        }



        public boolean hasNext() {

            

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

            if (st.size() > 0) {

                return true;

            }

            else {

                return false;

            }

        }



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

        private void leftMostInorder(TreeNode root) {

            while (root != null) {

                st.push(root);

                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 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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