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---
title: '102. Binary Tree Level Order Traversal'
description: 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)
sidebar:
label: 'Binary Tree Level Order Traversal'
badge: 'Medium'
---
<Badge variant="accent">Tree BFS</Badge>
### 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`
## Approach
```mermaid
flowchart TD
S(["levelOrder(root)"]) --> I["res = [], q = deque(), q.append(root)"]
I --> W{"q non-empty?"}
W -- no --> E(["return res"])
W -- yes --> L["qLen = len(q), level = [] — qLen freezes the current level's size"]
L --> F{"more i in range(qLen)?"}
F -- no --> V{"level non-empty?"}
V -- yes --> A["res.append(level)"]
A --> W
V -- no --> W
F -- yes --> P["node = q.popleft()"]
P --> N{"node is not None?"}
N -- no --> F
N -- yes --> C["level.append(node.val), q.append(node.left), q.append(node.right) — None children are filtered on the next level"]
C --> F
```
## Solution
```py
# 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
from collections import deque
class Solution:
# T: O(n) S: O(n)
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
res = []
q = deque()
q.append(root)
while q:
qLen = len(q)
level = []
for i in range(qLen):
node = q.popleft()
if node:
level.append(node.val)
q.append(node.left)
q.append(node.right)
if level:
res.append(level)
return res
```
## Explanation
<YouTube id="6ZnyEApgFYg" title="Binary Tree Level Order Traversal - BFS - Leetcode 102" />