Today we are going to solve HackerRank Day 23 : BST Level order traversal 30 days of code solution in C, C++, Java, Python & Javascript.
Objective
Today, we’re going further with Binary Search Trees.
Task
A level-order traversal, also known as a breadth-first search, visits each level of a tree’s nodes from left to right, top to bottom. You are given a pointer, root, pointing to the root of a binary search tree. Complete the levelOrder function provided in your editor so that it prints the level-order traversal of the binary search tree.
Hint: You’ll find a queue helpful in completing this challenge.
Function Description
Complete the levelOrder function in the editor below.
levelOrder has the following parameter:
– Node pointer root: a reference to the root of the tree
Prints
– Print node.data items as space-separated line of integers. No return value is expected.
Input Format
The locked stub code in your editor reads the following inputs and assembles them into a BST:
The first line contains an integer, T (the number of test cases).
The T subsequent lines each contain an integer, data, denoting the value of an element that must be added to the BST.
Constraints
- 1 <= N <= 20
- 1 <= node.data[i] <= 100
Output Format
Print the data value of each node in the tree’s level-order traversal as a single line of N space-separated integers.
Sample Input
6
3
5
4
7
2
1
Sample Output
3 2 5 1 4 7
Explanation
We traverse each level of the tree from the root downward, and we process the nodes at each level from left to right. The resulting level-order traversal is 3 = 2 = 5 = 1 = 4 = 7, and we print these data values as a single line of space-separated integers.
HackerRank Day 23 : BST Level order traversal 30 days of code solution
BST Level order traversal HackerRank Solution in C
#define max(a, b) (a > b ? a : b) int getHeight(Node *root) { if (root == NULL) return 0; else return 1 + max(getHeight(root->left), getHeight(root->right)); } void printGivenLevel(Node *root, int level) { if (root == NULL) return; if (level == 1) printf("%d ", root->data); else if (level > 1) { printGivenLevel(root->left, level-1); printGivenLevel(root->right, level-1); } } void levelOrder(Node* root){ //Write your code here int height = getHeight(root); int i; for (i = 1; i <= height; i++) { printGivenLevel(root, i); } }
BST Level order traversal HackerRank Solution in C++
void levelOrder(Node * root){ std::queue<Node*> q; Node* c; if (root != NULL) { q.push(root); } while (!q.empty()) { c = q.front(); q.pop(); cout << c->data << " "; if (c->left!=NULL) q.push(c->left); if (c->right!=NULL) q.push(c->right); } }
BST Level order traversal HackerRank Solution in Java
static LinkedList<Integer> queue = new LinkedList(); static void levelOrder(Node root){ LinkedList<Node> treeQueue = new LinkedList(); treeQueue.add(root); while(treeQueue.peek() != null) { Node toprint = treeQueue.remove(); System.out.print(toprint.data); if(toprint.left != null) { treeQueue.add(toprint.left); } if(toprint.right != null) { treeQueue.add(toprint.right); } if(treeQueue.peek() != null) { System.out.print(" "); } } }
Binary Search Trees HackerRank Solution in Python 3
def levelOrder(self,root): if root is None: return qu = [] qu.append(root) while len(qu) !=0: p = qu.pop(0) print(p.data, end=' ') if p.left is not None: qu.append(p.left) if p.right is not None: qu.append(p.right)
Binary Search Trees HackerRank Solution in JavaScript
var queue = [root]; while (queue.length > 0) { var node = queue.shift(); write(node.data + " "); if(node.left) { queue.push(node.left); } if (node.right) { queue.push(node.right); } } function write(str){ process.stdout.write(str); }
NEXT: HackerRank Day 24: More Linked Lists 30 days of code solution