49 lines
1.6 KiB
Python
49 lines
1.6 KiB
Python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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from collections import deque
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class Solution:
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def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
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'''
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BFS
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'''
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if root is None: return []
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q = deque([root])
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results = []
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while q:
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n = len(q)
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level = []
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# Idea here is we iterate through the queue, the queue
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# will always contain 1, 2, 4, 8 ... elements for each
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# level since we'll be appending None .left/.right nodes
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# also. This makes it easier to do the computation.
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for i in range(n):
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curr = q.popleft()
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# Could be None, ignore if that's the case
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if curr:
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level.append(curr.val)
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# It's okay to append None values here, the condition
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# above will filter it out
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q.append(curr.left)
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q.append(curr.right)
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# Since there's a chance that at the last level of the tree, we
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# might be pushing None nodes due to .left/.right being None, we
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# might have an empty level. We check and avoid adding the empty
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# level in this case.
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if level:
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results.append(level)
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return results
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