LeetCode 103: Binary Tree Zigzag Level Order Traversal

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In each iteration of BFS, we process entire level. For odd levels, order is reversed.

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 zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
            
        zigzag = []
        level_id = 0
        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)
            zigzag.append(level if level_id % 2 == 0 else level[::-1])
            level_id += 1

        return zigzag

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