## DEV Community

Abhishek Chaudhary

Posted on

# 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:

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`

SOLUTION:

``````# 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]]:
if not root:
return []
queue = [(root, 0)]
levels = {}
level = 0
while len(queue) > 0:
curr, level = queue.pop(0)
levels[level] = levels.get(level, []) + [curr.val]
if curr.left:
queue.append((curr.left, level + 1))
if curr.right:
queue.append((curr.right, level + 1))
return [levels[l] for l in range(level + 1)]
``````