BinaryTreeLevelOrderTraversal.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
"""
from collections import deque


# Definition for a binary tree node.
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: TreeNode) -> List[List[int]]:
        if not root:
            return []

        result = []
        queue = deque([root])

        while queue:
            level_size = len(queue)
            level_values = []

            for i in range(level_size):
                node = queue.popleft()
                level_values.append(node.val)

                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            result.append(level_values)

        return result


# Example 1
root1 = TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
solution = Solution()
print(solution.levelOrder(root1))  # Output should be [[3], [9, 20], [15, 7]]

# Example 2
root2 = TreeNode(1)
solution = Solution()
print(solution.levelOrder(root2))  # Output should be [[1]]

# Example 3
root3 = None
solution = Solution()
print(solution.levelOrder(root3))  # Output should be []