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