Updated task 102

Author: javadev

src/main/java/g0101_0200/s0102_binary_tree_level_order_traversal/readme.md

@@ -29,61 +29,60 @@ Given the `root` of a binary tree, return _the level order traversal of its node
 *   The number of nodes in the tree is in the range `[0, 2000]`.
 *   `-1000 <= Node.val <= 1000`
 
-To solve this problem in Java with a `Solution` class, we'll use a breadth-first search (BFS) traversal approach. Below are the steps:
+To solve the "Binary Tree Level Order Traversal" problem in Java with a `Solution` class, we'll perform a breadth-first search (BFS) traversal of the binary tree. Below are the steps:
 
 1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
 
 2. **Create a `levelOrder` method**: This method takes the root node of the binary tree as input and returns the level order traversal of its nodes' values.
 
-3. **Check for null root**: Check if the root is null. If so, return an empty list since there are no nodes to traverse.
+3. **Initialize a queue**: Create a queue to store the nodes during BFS traversal.
 
-4. **Initialize a queue**: We'll use a queue to perform BFS traversal. Add the root node to the queue.
+4. **Check for null root**: Check if the root is null. If it is, return an empty list.
 
-5. **Traverse the tree level by level**:
-   - Start a loop that continues until the queue is empty.
-   - Inside the loop, for each level, get the size of the queue (number of nodes at the current level).
-   - Create a list to store the node values at the current level.
-   - Iterate through all nodes at the current level:
-     - Poll a node from the queue.
-     - Add its value to the list.
-     - If the node has left or right children, add them to the queue.
-   - After processing all nodes at the current level, add the list of node values to the result list.
+5. **Perform BFS traversal**: Enqueue the root node into the queue. While the queue is not empty:
+   - Dequeue the front node from the queue.
+   - Add the value of the dequeued node to the current level list.
+   - Enqueue the left and right children of the dequeued node if they exist.
+   - Move to the next level when all nodes in the current level are processed.
 
-6. **Return the result**: After traversing the entire tree, return the result containing the level order traversal.
+6. **Return the result**: After the BFS traversal is complete, return the list containing the level order traversal of the binary tree.
 
 Here's the Java implementation:
 
 ```java
-import java.util.*;
+import java.util.ArrayList;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
 
 class Solution {
     public List<List<Integer>> levelOrder(TreeNode root) {
-        List<List<Integer>> result = new ArrayList<>();
-        
+        List<List<Integer>> result = new ArrayList<>(); // Initialize list to store level order traversal
         if (root == null) return result; // Check for empty tree
 
-        Queue<TreeNode> queue = new LinkedList<>();
-        queue.offer(root); // Add root to the queue
+        Queue<TreeNode> queue = new LinkedList<>(); // Initialize queue for BFS traversal
+        queue.offer(root); // Enqueue the root node
 
         while (!queue.isEmpty()) {
-            int levelSize = queue.size(); // Get the number of nodes at the current level
-            List<Integer> levelNodes = new ArrayList<>(); // List to store node values at current level
+            int levelSize = queue.size(); // Get the number of nodes in the current level
+            List<Integer> level = new ArrayList<>(); // Initialize list for the current level
 
             for (int i = 0; i < levelSize; i++) {
-                TreeNode node = queue.poll(); // Remove node from the queue
-                levelNodes.add(node.val); // Add its value to the list
+                TreeNode node = queue.poll(); // Dequeue the front node
+                level.add(node.val); // Add node value to the current level list
 
-                // Add left and right children to the queue if they exist
+                // Enqueue the left and right children if they exist
                 if (node.left != null) queue.offer(node.left);
                 if (node.right != null) queue.offer(node.right);
             }
-            result.add(levelNodes); // Add list of node values at current level to result
+            
+            result.add(level); // Add the current level list to the result list
         }
 
-        return result;
+        return result; // Return the level order traversal
     }
 
-    // TreeNode definition
+    // Definition for a TreeNode
     public class TreeNode {
         int val;
         TreeNode left;
@@ -100,4 +99,4 @@ class Solution {
 }
 ```
 
-This implementation follows the steps outlined above and efficiently solves the problem of binary tree level order traversal in Java.
+This implementation follows the steps outlined above and efficiently computes the level order traversal of the binary tree in Java using BFS.