Insert Into A Binary Search Tree
You have been given a root node of the binary search tree and a positive integer value. You need to perform an insertion operation i.e. inserting a new node with the given value in the given binary search tree such that the resultant tree is also a binary search tree. If there can be more than one possible tree, then you can return any.

Note :

A binary search tree is a binary tree data structure, with the following properties : a. The left subtree of any node contains nodes with the value less than the node’s value. b. The right subtree of any node contains nodes with the value equal to or greater than the node’s value. c. Right, and left subtrees are also binary search trees. It is guaranteed that, a. All nodes in the given tree are distinct positive integers. b. The given BST does not contain any node with a given integer value.

Example, below the tree, is a binary search tree.

Below the tree is not a BST as node ‘2’ is less than node ‘3’ but ‘2’ is the right child of ‘3’, and node ‘6’ is greater than node ‘5’ but it is in the left subtree of node ‘5’.

Input Format :

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. 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. The second line of each test case contains a positive integer value ‘val’, denoting the value of the node that is to be inserted in the given BST. For example, the level order input for the tree depicted in the below image.

50 13 72 3 25 66 -1 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 50 Level 2 : Left child of 50 = 13 Right child of 50 = 72 Level 3 : Left child of 13 = 3 Right child of 13 = 25 Left child of 72 = 66 Right child of 72 = ‘Null’ Level 4 : All children are ‘Null’ 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: 50 13 72 3 25 66 -1 -1 -1 -1 -1 -1 -1

Output Format :

For each test case, the output will be “1” if you have returned the correct answer, else it will be “0”. The output of each test case will be printed in a separate line.

Note :

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

Constraints :

1

Question

Insert Into A Binary Search Tree

You have been given a root node of the binary search tree and a positive integer value. You need to perform an insertion operation i.e. inserting a new node with the given value in the given binary search tree such that the resultant tree is also a binary search tree. If there can be more than one possible tree, then you can return any.

Note :

A binary search tree is a binary tree data structure, with the following properties :
    a. The left subtree of any node contains nodes with the value less than the node’s value.
    b. The right subtree of any node contains nodes with the value equal to or greater than the node’s value.
    c. Right, and left subtrees are also binary search trees.
It is guaranteed that,
    a. All nodes in the given tree are distinct positive integers.
    b. The given BST does not contain any node with a given integer value.

Example, below the tree, is a binary search tree.

Below the tree is not a BST as node ‘2’ is less than node ‘3’ but ‘2’ is the right child of ‘3’, and node ‘6’ is greater than node ‘5’ but it is in the left subtree of node ‘5’.

Input Format :

The first line contains an integer 'T' which denotes the number of test cases or queries to be run.

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.

The second line of each test case contains a positive integer value ‘val’, denoting the value of the node that is to be inserted in the given BST.

For example, the level order input for the tree depicted in the below image.

50 13 72 3 25 66 -1 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 50 Level 2 : Left child of 50 = 13 Right child of 50 = 72 Level 3 : Left child of 13 = 3 Right child of 13 = 25 Left child of 72 = 66 Right child of 72 = ‘Null’ Level 4 : All children are ‘Null’ 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: 50 13 72 3 25 66 -1 -1 -1 -1 -1 -1 -1

Output Format :

For each test case, the output will be “1” if you have returned the correct answer, else it will be “0”. The output of each test case will be printed in a separate line.

Note :

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

Constraints :

1 <= T <= 5 0 <= N <= 3000 1 <= data <= 10 ^ 9 1 <= val <= 10 ^ 9 Where ‘N’ is the total number of nodes in the given binary tree, ‘data’ is the value of the nodes of the given binary tree, and ‘val’ represents the value of node that is to be inserted in the given tree. For a single test case, all given ‘data’ and ‘val’ are distinct from each other. Time Limit: 1sec

Accepted Answer

Recursive Approach

The rules of the binary search tree say if any node value is less than the root node value, then this node must lie on the left subtree of the root else will lie on the right subtree. We can think of a recursive solution by breaking the task into a subtask, as any node can be either on the left subtree or right subtree of the root node. In BST we can find which subtree (left/right) to follow next by comparing the given value and root node’s value.

For example, if the given value is ‘3’ and the root node’s value is ‘5’ the node will be inserted in the left subtree of the root node as ‘3 < 5’. Similarly, if the given value is greater than the root node’s value we can assure that the node will insert on the right subtree of the root node. For a given tree we can find a required position by recursively calling the function to find the position in left subtree or right subtree after comparing values of nodes. When we reach a subtree whose root node is ‘Null’, then this means that we have reached the required position.

Algorithm :

Base case - If the root is NULL, return a tree node with a value same as the given value.
If the given value is greater than the value of the root node, then recursively call the function for the right subtree and update the right subtree with the returned tree from the recursive function.
Else, call the function for the left subtree and update the left subtree with the returned tree from the recursive function.
Return root node.

Space Complexity: O(n)Explanation:

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

In the worst case, the binary search tree can be linear, and the recursion stack will store function calls with all nodes in the tree with overall space complexity O(N).

Time Complexity: O(n)Explanation:

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

In the worst case, the binary search tree can be linear, and hence we need to visit every node with overall time complexity O(N)

Java (SE 1.8)

/*

    Time Complexity: O(N)

    Space Complexity: O(N)



    Where N is the total number of nodes in the given tree.

*/



import java.util.ArrayList;



public class Solution 

{

    // Recursive function to insert new node in correct position

    public static TreeNode insertionInBST(TreeNode root, int val) 

    {

        // If root is null, return a new node with given value.

        if (root == null)

        {

            TreeNode node = new TreeNode (val);

            return node;

        }



        /*

            If given value is less than root node's value, then node will insert in

            left subtree of root node Else it will insert in right subtree of root node.

        */



        if (root.val > val)

        {

            // Call recursive function with left child and update left subtree.

            root.left = insertionInBST(root.left, val);

        }



        else

        {

            // Call recursive function with right child and update right subtree.

            root.right = insertionInBST(root.right, val);

        }



        // Return updated tree

        return root;

    }

}

Python (3.5)

'''

    Time Complexity : O(N)

    Space Complexity: O(N)



    Where N is the total number of nodes in the given tree.

'''



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



# Recursive  function to insert new node in correct position

def insertionInBST(root,val):



    # If root is null, return a new node with given value.

    if (root == None):

        node = TreeNode(val)

        return node



    '''

        If given value is less than root node's value, then node will insert in 

        left subtree of root node Else it will insert in right subtree of root node.

    '''

    if (root.val > val):

        # Call recursive function with left child and update left subtree.

        root.left = insertionInBST(root.left, val)



    else:

        # Call recursive function with right child and update right subtree.

        root.right = insertionInBST(root.right, val)



    # Return updated tree

    return root

C++ (g++ 5.4)

/*

    Time Complexity: O(N)

    Space Complexity: O(N)



    Where N is the total number of nodes in the given tree.

*/



// Recursive  function to insert new node in correct position

TreeNode* insertionInBST(TreeNode* root, int val)

{

    // If root is null, return a new node with given value.

    if (root == NULL)

    {

        TreeNode* node = new TreeNode (val);

        return node;

    }



    /*

        If given value is less than root node's value, then node will insert in 

        left subtree of root node Else it will insert in right subtree of root node.

    */



    if (root->val > val)

    {

        // Call recursive function with left child and update left subtree.

        root->left = insertionInBST(root->left, val);

    }



    else

    { 

        // Call recursive function with right child and update right subtree.

        root->right = insertionInBST(root->right, val);

    }



    // Return updated tree

    return root;

}

Accepted Answer

Iterative Approach

In order to find a correct position for the insertion of a given node, we can find a path from the root node to a node whose left/right child can be the node that is to be inserted. We can follow from root to down the tree by choosing the left subtree or the right subtree by comparing the value of the root node of the subtree with the given value.

Algorithm :

If the root is NULL
- Return new node with value same as the given value.
Declare a variable node say ‘temp’ and initialize it with the root node.
Run a loop until ‘temp’ is not NULL.
- If the given value is greater than the ‘temp -> val’
  - Set temp = temp -> right i.e. ( move to right subtree ).
- Else
  - Set temp = temp -> left i.e. ( move to left subtree ).
We have reached the required position, So set temp to the new node with the given value.
Return the ‘root node’.

Space Complexity: O(1)Explanation:

O(1).

We are not using any extra space in this algorithm, Hence the space complexity is Constant.

Time Complexity: O(n)Explanation:

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

In the worst case, when the given tree is linear, we have to visit every node exactly once Hence, the overall time complexity will be O(N).

Java (SE 1.8)

/*

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where N is the total number of nodes in given binary tree.

*/



import java.util.ArrayList;



public class Solution 

{

    public static TreeNode insertionInBST(TreeNode root, int val) 

	{

        // If root is null, return a new node with the given value.

        if (root == null)

        {

            TreeNode node = new TreeNode (val);

            return node;

        }



        // Declare a 'temp' node and initialize with the given root node.

        TreeNode temp = root;



        // Create a new 'node' that is to be inserted in the given tree

        TreeNode node = new TreeNode (val);



        // Run a loop until 'temp' is not null.

        while (temp != null)

        {

            // If given value is less than temp -> val, 'node' will be inserted in left subtree.

            if (temp.val > val)

            {

                // If left child of temp is null, set left child to 'node'.

                if (temp.left == null)

                {

                    temp.left = node;

                    break;

                }



                temp = temp.left;

            }



            // If given value is greator than temp -> val, 'node' will be inserted in right subtree.

            else

            {

                // If right child is NULL, set right child to 'node'.

                if (temp.right == null)

                {

                    temp.right = node;

                    break;

                }



                temp = temp.right;

            }

        }



        // Return updated tree

        return root;

    }

}

Python (3.5)

'''

    Time Complexity : O(N)

    Space Complexity: O(1)



    Where N is the total number of nodes in given binary tree.

'''



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



def insertionInBST(root,val):



    # If root is null, return a new node with the given value.

    if (root == None):

        node = TreeNode(val);

        return node



    # Declare a 'temp' node and initialize with the given root node.

    temp = root



    # Create a new 'node' that is to be inserted in the given tree

    node = TreeNode(val)



    # Run a loop until 'temp' is not null.

    while (temp != None):



        # If given value is less than temp.val, 'node' will be inserted in left subtree. 

        if (temp.val > val):



            # If left child of temp is null, set left child to 'node'.

            if(temp.left == None):

                temp.left = node

                break



            temp = temp.left



        # If given value is greator than temp.val, 'node' will be inserted in right subtree. 

        else :

            

            # If right child is None, set right child to 'node'.

            if (temp.right == None):

                temp.right = node

                break



            temp = temp.right



    # Return updated tree

    return root

C++ (g++ 5.4)

/*

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where N is the total number of nodes in given binary tree.

*/



TreeNode* insertionInBST(TreeNode* root, int val)

{

    // If root is null, return a new node with the given value.

    if (root == NULL)

    {

        TreeNode* node = new TreeNode (val);

        return node;

    }



    // Declare a 'temp' node and initialize with the given root node.

    TreeNode* temp = root;



    // Create a new 'node' that is to be inserted in the given tree

    TreeNode* node = new TreeNode (val);



    // Run a loop until 'temp' is not null.

    while (temp != NULL)

    {

        // If given value is less than temp -> val, 'node' will be inserted in left subtree. 

        if (temp -> val > val)

        {

            // If left child of temp is null, set left child to 'node'.

            if(temp -> left == NULL )

            {

                temp -> left = node;

                break;

            }



            temp = temp -> left;

        } 



        // If given value is greator than temp -> val, 'node' will be inserted in right subtree. 

        else 

        {

            // If right child is NULL, set right child to 'node'.

            if (temp -> right == NULL)

            {

                temp -> right = node;

                break;

            }



            temp = temp -> right;

        }

    }



    // Return updated tree

    return root;

}

Note :

Example, below the tree, is a binary search tree.

Below the tree is not a BST as node ‘2’ is less than node ‘3’ but ‘2’ is the right child of ‘3’, and node ‘6’ is greater than node ‘5’ but it is in the left subtree of node ‘5’.

Input Format :

Explanation :

Note :

Output Format :

Note :

Constraints :

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