LeetCode 102: Binary Tree Level Order Traversal

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In each iteration of BFS, we process the entire level.

Time: \mathcal{O}(n), space: \mathcal{O}(n).

# 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
class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        all_levels = []
        q = deque([root])
        while q:
            level_len = len(q)
            level = [0] * level_len
            for i in range(level_len):
                u = q.popleft()
                level[i] = u.val
                q.extend(c for c in (u.left, u.right) if c)
            all_levels.append(level)

        return all_levels

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