#
# @lc app=leetcode id=102 lang=python3
#
# [102] Binary Tree Level Order Traversal
#
# https://leetcode.com/problems/binary-tree-level-order-traversal/description/
#
# algorithms
# Medium (56.71%)
# Likes:    4342
# Dislikes: 103
# Total Accepted:    790K
# Total Submissions: 1.4M
# Testcase Example:  '[3,9,20,null,null,15,7]'
#
# 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
# 
# 
#

# @lc code=start
# 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]]:
        result = []

        def traverse(root):
            if root:
                result.append()

            return root.left.val 



# @lc code=end


#
# @lc app=leetcode id=102 lang=python3
#
# [102] Binary Tree Level Order Traversal
#
# https://leetcode.com/problems/binary-tree-level-order-traversal/description/
#
# algorithms
# Medium (56.71%)
# Likes:    4342
# Dislikes: 103
# Total Accepted:    790K
# Total Submissions: 1.4M
# Testcase Example:  '[3,9,20,null,null,15,7]'
#
# 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
# 
# 
#

# @lc code=start
# 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]]:
        result = []

        def traverse(root):
            if root:
                result.append()

            return root.left.val 



# @lc code=end


# # @lc app=leetcode id=102 lang=python3 # # [102] Binary Tree Level Order Traversal # # https://leetcode.com/problems/binary-tree-level-order-traversal/description/ # # algorithms # Medium (56.71%) # Likes:    4342 # Dislikes: 103 # Total Accepted:    790K # Total Submissions: 1.4M # Testcase Example:  '[3,9,20,null,null,15,7]' # # 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 # # # # @lc code=start # 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 # @lc code=end