# 102 Binary Tree Level Order Traversal

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:

``````  3
/ \
9  20
/ \
15  7

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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 `````` ``````# 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: Optional[TreeNode]) -> List[List[int]]: def traverse(root, level, result): if not root: return if level > len(result)-1: result.append([]) result[level].append(root.val) traverse(root.left, level+1, result) traverse(root.right, level+1, result) result = [] traverse(root, 0, result) return result``````