题目链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
难度：中等

描述：
给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。

题解

  1. from collections import deque
  2. # Definition for a binary tree node.
  3. # class TreeNode:
  4. #     def __init__(self, val=0, left=None, right=None):
  5. #         self.val = val
  6. #         self.left = left
  7. #         self.right = right
  8. class Solution:
  9.     def levelOrder(self, root: TreeNode) -> List[List[int]]:
  10.         ret = []
  11.         if not root:
  12.             return ret
  14.         q = deque()
  15.         q.append(root)
  16.         while q:
  17.             size = len(q)
  18.             cur_level = []
  19.             while size > 0:
  20.                 temp_node = q.popleft()
  21.                 cur_level.append(temp_node.val)
  22.                 if temp_node.left is not None:
  23.                     q.append(temp_node.left)
  24.                 if temp_node.right is not None:
  25.                     q.append(temp_node.right)
  26.                 size -= 1
  27.             ret.append(cur_level)
  28.         return ret
  1. # Definition for a binary tree node.
  2. # class TreeNode:
  3. #     def __init__(self, val=0, left=None, right=None):
  4. #         self.val = val
  5. #         self.left = left
  6. #         self.right = right
  7. class Solution:
  8.     def levelOrder(self, root: TreeNode) -> List[List[int]]:
  9.         ret = []
  11.         if not root:
  12.             return ret
  14.         def dfs(node, level):
  15.             if node is None:
  16.                 return
  18.             if level >= len(ret):
  19.                 ret.append([])
  20.             ret[level].append(node.val)
  22.             if node.left is not None:
  23.                 dfs(node.left, level+1)
  24.             if node.right is not None:
  25.                 dfs(node.right, level+1)
  26.         dfs(root, 0)
  27.         return ret