The code must be done in C++. Please read the question very carefully and write just according to the question.I repet exactly according to the question.

1. [26 points]

a) Write a function that will print the even elements in each level of the binary search tree.

b) Write a function that will count total number of odd nodes of the tree.

10

/ \

6 13

/ \ / \

3 8 11 16

The output of the given tree when the function is applied:

10

6

8 16

Total number of odd nodes is 3.

You have to answer both (a) and (b) in same code calling from main function. The base code is given bellow. Modify and write only the main function and the additional function needed to solve the problem in the answer script.

BST code:

#include

using namespace std;

struct node

{

int key;

struct node *left, *right;

};

// A utility function to search a given node

bool search(struct node* root, int key)

{

// Base Cases: root is null or key is present at root

if (root == NULL)

return false;

else if(root->key == key)

return true;

// Key is greater than root's key

else if (root->key < key)

return search(root->right, key);

else

// Key is smaller than root's key

return search(root->left, key);

}

// A utility function to create a new BST node

struct node *newNode(int item)

{

struct node *temp = new node;

temp->key = item;

temp->left = temp->right = NULL;

return temp;

}

// A utility function to do inorder traversal of BST

void inorder(struct node *root)

{

if (root != NULL)

{

inorder(root->left);

cout<key<<" ";

inorder(root->right);

}

}

// A utility function to insert a new node with given key in BST

struct node* insert(struct node* node, int key)

{

/* If the tree is empty, return a new node */

if (node == NULL) return newNode(key);

/* Otherwise, recur down the tree */

if (key < node->key)

node->left = insert(node->left, key);

else if (key > node->key)

node->right = insert(node->right, key);

/* return the (unchanged) node pointer */

return node;

}

// Driver Program to test above functions

int main()

{

struct node *root = NULL;

root = insert(root, 50);

insert(root, 30);

insert(root, 20);

insert(root, 40);

insert(root, 70);

insert(root, 60);

insert(root, 80);

if(search(root,60)==1)

cout<<"Node Present ";

else

cout<<"Node not present ";

// print inoder traversal of the BST

inorder(root);

return 0;

}