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- 30 Days of Code
- Day 23: BST Level-Order Traversal

# Day 23: BST Level-Order Traversal

# Day 23: BST Level-Order Traversal

- Prepare
- Tutorials
- 30 Days of Code
- Day 23: BST Level-Order Traversal

**Objective**

Today, we're going further with Binary Search Trees. Check out the Tutorial tab for learning materials and an instructional video!

**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, , 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, (the number of test cases).

The subsequent lines each contain an integer, , denoting the value of an element that must be added to the BST.

**Constraints**

**Output Format**

Print the value of each node in the tree's level-order traversal as a single line of space-separated integers.

**Sample Input**

```
6
3
5
4
7
2
1
```

**Sample Output**

```
3 2 5 1 4 7
```

**Explanation**

The input forms the following binary search tree:

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 , and we print these data values as a single line of space-separated integers.