LeetCode-in-Java

102. Binary Tree Level Order Traversal

Medium

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: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

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. Initialize a queue: Create a queue to store the nodes during BFS traversal.

4. Check for null root: Check if the root is null. If it is, return an empty 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 the BFS traversal is complete, return the list containing the level order traversal of the binary tree.

Here’s the Java implementation:

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<>(); // Initialize list to store level order traversal
        if (root == null) return result; // Check for empty tree
        
        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 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(); // Dequeue the front node
                level.add(node.val); // Add node value to the current level list
                
                // 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(level); // Add the current level list to the result list
        }
        
        return result; // Return the level order traversal
    }
    
    // Definition for a TreeNode
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        
        TreeNode() {}
        TreeNode(int val) { this.val = val; }
        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }
}

This implementation follows the steps outlined above and efficiently computes the level order traversal of the binary tree in Java using BFS.

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