标签：put continue none obj pap ret lazy turn man
Description: Given the root of a binary tree, return the bottom-up level order traversal of its nodes‘ values. (i.e., from left to right, level by level from leaf to root).
Link: 107. Binary Tree Level Order Traversal II
Exaples:
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[15,7],[9,20],[3]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: []
思路: 因为是按照每一层返回，所以用广搜，BFS每一层的val，然后reverse list返回。
class Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if root is None: return [] res = [] que = collections.deque() que.append(root) while que: size = len(que) layerval = [] for _ in range(size): node = que.popleft() if node is None: continue layerval.append(node.val) que.append(node.left) que.append(node.right) if len(layerval)>0: res.append(layerval) return res[::-1]
日期: 2021-03-13 (Peer presure is so heavy, many papers are accepted in our group by NAACL, feel so stressed.)
Leetcode 107. Binary Tree Level Order Traversal II
标签：put continue none obj pap ret lazy turn man
原文地址：https://www.cnblogs.com/wangyuxia/p/14528887.html
