Introduction to the Problem

Binary trees are a fundamental data structure in computer science, used in various applications ranging from data organization to decision processes.

A common task in software engineering interviews, including platforms like LeetCode, involves traversing these trees in a level order manner (LeetCode 102). This means processing the nodes of the tree level by level from left to right, which can be crucial for understanding the structure and logic of a binary tree.

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Consideration of Different Approaches

When tasked with performing a level order traversal of a binary tree, the most straightforward and efficient approach is using Breadth-First Search (BFS). This method uses a queue to keep track of nodes and process them level by level.

An alternative approach is Depth-First Search (DFS), which, although useful for many tree-related problems, is less efficient for level order traversal. DFS would require additional mechanisms to track node levels, making the solution more complex and less intuitive than BFS for this specific task.

The Solution

• To solve the Binary Tree Level Order Traversal, we initiate a queue that starts with the root node.

• While the queue is not empty, we remove a node from the front, process it, and then add its children (first left, then right) to the back of the queue.

• This continues until the queue is empty, ensuring that nodes are processed in a level-wise sequence.

• The complexity of this solution is O(n), where n is the number of nodes in the tree, as each node is processed exactly once.

The Solution in Python

Here’s how you can implement the BFS for a binary tree level order traversal in Python:

def levelOrder(root):
    if not root:
        return []

    from collections import deque
    queue = deque([root])
    result = []

    while queue:
        level_size = len(queue)
        level_values = []

        for _ in range(level_size):
            node = queue.popleft()
            level_values.append(node.val)

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        result.append(level_values)

    return result

The Solution in TypeScript

Implementing the same logic in TypeScript with added type safety:

function levelOrder(root: TreeNode | null): number[][] {
    if (!root) return [];

    const result: number[][] = [];
    const queue: TreeNode[] = [root];

    while (queue.length > 0) {
        const levelSize: number = queue.length;
        const level: number[] = [];

        for (let i = 0; i < levelSize; i++) {
            const currentNode = queue.shift();

            if (currentNode) {
                level.push(currentNode.val);

                if (currentNode.left) {
                    queue.push(currentNode.left);
                }
                if (currentNode.right) {
                    queue.push(currentNode.right);
                }
            }
        }

        result.push(level);
    }

    return result;
}

The Solution in Java

Finally, a Java implementation which adheres closely to the same principles:

import java.util.List;
import java.util.ArrayList;
import java.util.Queue;
import java.util.LinkedList;

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            List<Integer> level = new ArrayList<>();

            for (int i = 0; i < levelSize; i++) {
                TreeNode currentNode = queue.poll();
                level.add(currentNode.val);

                if (currentNode.left != null) {
                    queue.offer(currentNode.left);
                }
                if (currentNode.right != null) {
                    queue.offer(currentNode.right);
                }
            }
            result.add(level);
        }



        return result;
    }
}

Conclusion

Understanding and implementing the Binary Tree Level Order Traversal is a valuable skill, particularly for LeetCode-style interviews. By using a BFS approach, you ensure that the solution is both efficient and straightforward, handling each node exactly once.

This exploration of the problem in three different programming languages not only highlights the universal applicability of the BFS method but also prepares you for handling similar problems in various technical interviews.