Write a C++ program that creates a tree from given input, and prints the leaf no
ID: 3684355 • Letter: W
Question
Write a C++ program that creates a tree from given input, and prints the leaf nodes using
the level-order traversal.
The following is how we define tree nodes:
struct node
{ int item; node *l, *r;
node(int x)
{ item = x; l = 0; r = 0; }
};
typedef node* link;
The trees for this assignment have the following properties:
1) Each node has zero, one, or two children (left child and right child);
2) Each node has a unique key. That is, the item in each node is different from items
in all other nodes.
3) Each input item is also unique.
You may find the following function useful. This function visits nodes in a tree in level
order starting at the given root h using a QUEUE (you can use any implementation for
QUEUE):
void traverse(link root, int n)
{
link h;
QUEUE q(n);
q.put(root);
while (!q.empty())
{
h = q.get(); // get link h at the head of the queue
if (h->l != 0) q.put(h->l);
if (h->r != 0) q.put(h->r);
}
}
Your program will receive the following input and do the following for each input:
C <int r> : Create the root node with item r. If the tree is not empty (i.e. the root exists),
return an error
1
L <int p> <int l>: Find the node with item p. Let u be that node. Create a new node v
with item l. Make node v node u’s left child. If node u has already a left child, return an
error
R <int p> <int r>: Find the node with item p. Let u be that node. Create a new node v
with item r. Make node v node u’s right child. If node u has already a right child, return
an error
P: Print the items in nodes in a level order traversal manner
S <int p>: Find the node with item p. Let u be that node. Print the items in nodes in the
subtree rooted at u in level order traversal manner
See the example dialogue below:
C 5
Root node with item 5 has been created
C 10
Error: Tree is not empty
L 5 3
Node with item 3 has been added
R 5 4
Node with item 4 has been added
P
level order traversal of the entire tree: 5 3 4
S 4
Level order traversal of the asked subtree: 4
L 4 7
R 7 8
R 3 9
L 9 11
R 4 12
L 7 13
S 3
level order traversal of the asked subtree: 3 9 11
S 4
level order traversal of the asked subtree: 4 7 12 13 8
L 8 14
R 12 15
R 8 16
L 13 17
P
level order traversal of the entire tree: 5 3 4 9 7 12 11 13 8 15 17 14 16
Explanation / Answer
// This is the Binary Tree Node to create a binary tree
struct Node
{
int data;
struct Node *left, *right;
};
// This is an iterative method to find height of Binary Tree
void toprintLevelOrdertree(Node *root)
{
//This will be the base Case to see if the tree is empty
if (root == NULL) return;
// now we will create an empty queue for level order tarversal
queue<Node *> z;
// now we will be enqueuing Root and initialize the height
z.push(root);
while (z.empty() == false)
{
// Next we will print front of queue and remove it from queue
Node *node = z.front();
cout << node->data << " ";
z.pop();
/* now we will Enqueue the left child */
if (node->left != NULL)
z.push(node->left);
/*Finally we will enqueue right child of the tree */
if (node->right != NULL)
z.push(node->right);
}
}
// This is a function to create a new tree node here
Node* createnewNode(int data)
{
Node *temporary = new Node;
temporary->data = data;
temporary->left = temporary->right = NULL;
return temporary;
}
//This is the given driver program to test above functions
int main()
{
//This is used to create a binary tree
Node *root = createnewNode(5);
root->left = createnewNode(6);
root->right = createnewNode(7);
root->left->left = createnewNode(8);
root->left->right = createnewNode(10);
cout << "This is the given Level Order traversal of binary tree is ";
toprintLevelOrdertree(root);
return 0;
}
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