leetcode-102 Binary Tree Level Order Traversal

# 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

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

from collections import deque

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []

        result = []
        queue = deque([root])

        while queue:
            level_size = len(queue)  # Number of nodes at current level
            current_level = []

            # Process all nodes at current level
            for _ in range(level_size):
                node = queue.popleft()
                current_level.append(node.val)

                # Add children to queue for next level
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            result.append(current_level)

        return result

# Time Complexity: O(n) - visit each node once
# Space Complexity: O(w) - where w is maximum width of tree (queue size)
#
# This BFS pattern is fundamental for many tree problems:
# - Level order traversal variants
# - Zigzag traversal
# - Right side view
# - Average of levels
# Remember: len(queue) before processing gives exact level size!