[LeetCode] 152. Binary Tree Upside Down 二叉树的上下颠倒
Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.
Example:
Input: [1,2,3,4,5]
1
/ \
2 3
/ \
4 5
Output: return the root of the binary tree [4,5,2,#,#,3,1]
4
/ \
5 2
/ \
3 1
Clarification:
Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].
这道题让我们把一棵二叉树上下颠倒一下,而且限制了右节点要么为空要么一定会有对应的左节点。上下颠倒后原来二叉树的最左子节点变成了根节点,其对应的右节点变成了其左子节点,其父节点变成了其右子节点,相当于顺时针旋转了一下。对于一般树的题都会有迭代和递归两种解法,这道题也不例外,先来看看递归的解法。对于一个根节点来说,目标是将其左子节点变为根节点,右子节点变为左子节点,原根节点变为右子节点,首先判断这个根节点是否存在,且其有没有左子节点,如果不满足这两个条件的话,直接返回即可,不需要翻转操作。那么不停的对左子节点调用递归函数,直到到达最左子节点开始翻转,翻转好最左子节点后,开始回到上一个左子节点继续翻转即可,直至翻转完整棵树,参见代码如下:
解法一:
class Solution {
public:
TreeNode *upsideDownBinaryTree(TreeNode *root) {
if (!root || !root->left) return root;
TreeNode *l = root->left, *r = root->right;
TreeNode *res = upsideDownBinaryTree(l);
l->left = r;
l->right = root;
root->left = NULL;
root->right = NULL;
return res;
}
};
下面我们来看迭代的方法,和递归方法相反的时,这个是从上往下开始翻转,直至翻转到最左子节点,参见代码如下:
解法二:
class Solution {
public:
TreeNode *upsideDownBinaryTree(TreeNode *root) {
TreeNode *cur = root, *pre = NULL, *next = NULL, *tmp = NULL;
while (cur) {
next = cur->left;
cur->left = tmp;
tmp = cur->right;
cur->right = pre;
pre = cur;
cur = next;
}
return pre;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/156
类似题目:
参考资料:
https://leetcode.com/problems/binary-tree-upside-down/
https://leetcode.com/problems/binary-tree-upside-down/discuss/49412/Clean-Java-solution
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] 152. Binary Tree Upside Down 二叉树的上下颠倒的更多相关文章
- [LeetCode] 156. Binary Tree Upside Down 二叉树的上下颠倒
Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that ...
- [LeetCode] Binary Tree Upside Down 二叉树的上下颠倒
Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that ...
- [LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tr ...
- ✡ leetcode 156. Binary Tree Upside Down 旋转树 --------- java
156. Binary Tree Upside Down Add to List QuestionEditorial Solution My Submissions Total Accepted: ...
- 【LeetCode】Binary Tree Upside Down
Binary Tree Upside Down Given a binary tree where all the right nodes are either leaf nodes with a s ...
- [leetcode]156.Binary Tree Upside Down颠倒二叉树
Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that ...
- LeetCode 563. Binary Tree Tilt (二叉树的倾斜度)
Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the ab ...
- LeetCode 257. Binary Tree Paths (二叉树路径)
Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 ...
- C++版 - LeetCode 144. Binary Tree Preorder Traversal (二叉树先根序遍历,非递归)
144. Binary Tree Preorder Traversal Difficulty: Medium Given a binary tree, return the preorder trav ...
随机推荐
- Autoware 笔记 No. 5——基于GNSS的定位
1. 前言 在之前的笔记No.2 中,我们直接采用ndt_matching的方法实现定位,但需要在打开rviz中,通过2D Pose Estimate指定初始位置.加入GNSS后,可以帮助ndt_ma ...
- 关于Idea突然无法输入的诡异问题解决
问题描述 最近加班把自己的装有Debian的笔记本带到公司,使用Idea写代码的时候,突然间无法输入,ctrl与tab还可用,重启Idea能得到一阵的解决 解决参考 如果是Linux平台,请考虑是否是 ...
- Lucene搜索/索引过程笔记
lucene索引文档过程: > 初始化IndexWriter > 构建Document > 调用IndexWriter.addDocument执行写入 > 初始化Documen ...
- 个人项目wc(Java)
个人项目(Java) 一丶Github地址:https://github.com/SAH2019/S ...
- 白话SCRUM之五:四种会议
在SCRUM方法中定义了4种会议活动: Sprint planning Daily meeting Sprint review Sprint retrospective 除去开发活动外这4种会议构成了 ...
- 简单的基于promise的ajax封装
基于promise的ajax封装 //调用方式: /* ajaxPrmomise({ url:, method:, headers:{} }).then(res=>{}) */ ;(functi ...
- Java后端面经总结:拿下蚂蚁金服美团头条 offer 秘诀
笔者在面过 猿辅导,去哪儿,旷视, 陌陌,头条, 阿里, 快手, 美团, 腾讯之后,除了收获一大堆面试问题,还思考到如何成为面试官眼中的”爱技术,爱思考,靠谱,有潜力候选人的”一些”套路”. 面试问题 ...
- E06 【买衣服】Maybe you need a bigger size
核心句型 Maybe you need a bigger size 也许您需要大一些的. 场景对话 A:Can I try this jacket on,please? 我能试试这件夹克吗? B:Su ...
- c# 第七节 编程规范,vs中的各种设置
本节内容: 1:编程规范 2:vs中的各种设置 3:一个解决方案多个项目的创建.使用 1:编程规范 需要规范的地方: 1: 2: 2:vs中的各种设置 实现: 3:一个解决方案多个项目的创建.使用
- Linux用户态与内核态通信的几种方式
本文首发于我的公众号 Linux云计算网络(id: cloud_dev),专注于干货分享,号内有 10T 书籍和视频资源,后台回复「1024」即可领取,欢迎大家关注,二维码文末可以扫. Linux 用 ...