[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. (i.e., 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

內容目錄

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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