102.py

'''
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
'''

# https://leetcode.com/problems/binary-tree-level-order-traversal/

import unittest

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def levelOrder(root):
    if not root:
        return []
    result = []
    queue = [root]
    while queue:
        level = []
        for _ in range(len(queue)):
            node = queue.pop(0)
            level.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        result.append(level)
    return result

class TestLevelOrder(unittest.TestCase):
    def test1(self):
        root = TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
        self.assertEqual(levelOrder(root), [[3], [9, 20], [15, 7]])

    def test2(self):
        root = TreeNode(1)
        self.assertEqual(levelOrder(root), [[1]])

    def test3(self):
        root = None
        self.assertEqual(levelOrder(root), [])

if __name__ == '__main__':
    unittest.main()