LeetCode 102 Binary Tree Level Order Traversal

LeetCode

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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:

<pre>Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] </pre>

Example 2:

<pre>Input: root = [1] Output: [[1]] </pre>

Example 3:

<pre>Input: root = [] Output: [] </pre>

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

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题目大意：

二叉树按层遍历

解题思路：

用BFS模板

解题步骤：

N/A

注意事项：

1. res.append(level)不是res.append(list(level))因为level = []已重新初始化。
2. 关键行： 多这一行level.append(node.val)，其他一样

Python代码：

def levelOrder(self, root: TreeNode) -> List[List[int]]:
	if not root:
		return []
	queue, res = collections.deque([root]), []
	while queue:
		level = []
		for _ in range(len(queue)):
			node = queue.popleft()
			level.append(node.val)

			if node.left:
				queue.append(node.left)
			if node.right:
				queue.append(node.right)
		res.append(level)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)