[Leetcode] 0102. Binary Tree Level Order Traversal

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Given the root of a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

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

内容目录

Python

				
					# time complexity: O(n) # space complexity: O(n) from typing import List, Optional 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]]: result: List[List[int]] = [] def bfs(node: Optional[TreeNode], level: int): if node is None: return if len(result) == level: result.append([]) result[level].append(node.val) if node.left: bfs(node.left, level + 1) if node.right: bfs(node.right, level + 1) bfs(root, 0) return result root = TreeNode(3) root.left = TreeNode(9) root.right = TreeNode(20) root.right.left = TreeNode(15) root.right.right = TreeNode(7) print(Solution().levelOrder(root))

Python

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