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
# 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]]:
queue, output = [(root, 1)], []
while queue:
item, level = queue.pop(0)
#item, level = data[0], data[1]
if item:
if len(output) < level:
output.append([item.val])
else:
output[level-1].append(item.val)
if item.left:
queue.append((item.left, level+1))
if item.right:
queue.append((item.right, level+1))
return output