leetcode: 107. Binary Tree Level Order Traversal II

最新推荐文章于 2020-03-09 16:24:05 发布

JNingWei

最新推荐文章于 2020-03-09 16:24:05 发布

阅读量204

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode-Python

本文链接：https://blog.youkuaiyun.com/JNingWei/article/details/83752723

LeetCode-Python 专栏收录该内容

134 篇文章

订阅专栏

本文探讨了如何使用广度优先搜索（BFS）策略实现二叉树的底层至上层的遍历，从叶节点开始返回每一层节点的值。通过实例演示了算法的具体实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Difficulty

Easy.

Problem

Given a binary tree, return the bottom-up level order traversal of its nodes' values. 
(ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

AC

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


# BFS
class Solution():
    def levelOrderBottom(self, root):
        res = []
        if root:
            level = [root]
            while len(level):
                res = [[node.val for node in level]] + res
                tmp = []
                for node in level:
                    if node.left:
                        tmp.append(node.left)
                    if node.right:
                        tmp.append(node.right)
                level = tmp
        return res