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