Table of Contents

## Question

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`

## Python 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: TreeNode) -> List[List[int]]: if root==None:return [] q = [root] res = [] while q: qtemp = [] r = [] while q: x = q.pop(0) r.append(x.val) if x.left:qtemp.append(x.left) if x.right:qtemp.append(x.right) res.append(r) q = qtemp return res