[LeetCode] Construct String from Binary Tree 根据二叉树创建字符串
You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.
The null node needs to be represented by empty parenthesis pair "()". And you need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.
Example 1:
Input: Binary tree: [1,2,3,4]
1
/ \
2 3
/
4 Output: "1(2(4))(3)"
Explanation: Originallay it needs to be "1(2(4)())(3()())",
but you need to omit all the unnecessary empty parenthesis pairs.
And it will be "1(2(4))(3)".
Example 2:
Input: Binary tree: [1,2,3,null,4]
1
/ \
2 3
\
4 Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example,
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
这道题给我们了一个二叉树,让我们创建对应的字符串,之前有一道正好反过来的题Construct Binary Tree from String。对于二叉树的处理,递归肯定是王道啊。想想如何来实现递归函数,我们观察到题目中的例子,发现如果左子结点为空,右子结点不为空时,需要在父结点后加上个空括号,而右子结点如果不存在,或者左右子结点都不存在就不需要这么做。那我们在递归函数中,如果当前结点不存在,直接返回,然后要在当前结点值前面加上左括号,然后判断,如果左子结点不存在,而右子结点存在的话,要在结果res后加上个空括号,然后分别对左右子结点调用递归函数,调用完之后要加上右括号,形成封闭的括号。由于最外面一层的括号不需要,所以我们再返回最终结果之前要去掉首尾的括号,参见代码如下:
class Solution {
public:
string tree2str(TreeNode* t) {
if (!t) return "";
string res = "";
helper(t, res);
return string(res.begin() + , res.end() - );
}
void helper(TreeNode* t, string& res) {
if (!t) return;
res += "(" + to_string(t->val);
if (!t->left && t->right) res += "()";
helper(t->left, res);
helper(t->right, res);
res += ")";
}
};
下面来看一种不用额外函数的递归写法,这种做法是一开始调用递归函数求出左右子结点的返回字符串,如果左右结果串均为空,则直接返回当前结点值;如果左子结果串为空,那么返回当前结果res,加上一个空括号,再加上放在括号中的右子结果串;如果右子结果串为空,那么发返回当前结果res,加上放在括号中的左子结果串;如果左右子结果串都存在,那么返回当前结果,加上分别放在括号中的左右子结果串,参见代码如下:
解法二:
class Solution {
public:
string tree2str(TreeNode* t) {
if (!t) return "";
string res = to_string(t->val);
string left = tree2str(t->left), right = tree2str(t->right);
if (left == "" && right == "") return res;
if (left == "") return res + "()" + "(" + right + ")";
if (right == "") return res + "(" + left + ")";
return res + "(" + left + ")" + "(" + right + ")";
}
};
下面这种解法更加简洁,由热心网友edyyy提供,思路和上面解法相同,参见代码如下:
解法三:
class Solution {
public:
string tree2str(TreeNode* t) {
if (!t) return "";
string res = to_string(t->val);
if (!t->left && !t->right) return res;
res += "(" + tree2str(t->left) + ")";
if (t->right) res += "(" + tree2str(t->right) + ")";
return res;
}
};
类似题目:
Construct Binary Tree from String
参考资料:
https://discuss.leetcode.com/topic/91308/java-solution-tree-traversal
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Construct String from Binary Tree 根据二叉树创建字符串的更多相关文章
- LeetCode 606. Construct String from Binary Tree根据二叉树创建字符串 (C++)
题目: You need to construct a string consists of parenthesis and integers from a binary tree with the ...
- Leetcode606.Construct String from Binary Tree根据二叉树创建字符串
你需要采用前序遍历的方式,将一个二叉树转换成一个由括号和整数组成的字符串. 空节点则用一对空括号 "()" 表示.而且你需要省略所有不影响字符串与原始二叉树之间的一对一映射关系的空 ...
- LeetCode Construct String from Binary Tree
原题链接在这里:https://leetcode.com/problems/construct-string-from-binary-tree/#/description 题目: You need t ...
- 606. Construct String from Binary Tree 从二叉树中构建字符串
[抄题]: You need to construct a string consists of parenthesis and integers from a binary tree with th ...
- 606. Construct String from Binary Tree 【easy】
606. Construct String from Binary Tree [easy] You need to construct a string consists of parenthesis ...
- 【Leetcode_easy】606. Construct String from Binary Tree
problem 606. Construct String from Binary Tree 参考 1. Leetcode_easy_606. Construct String from Binary ...
- LeetCode 606. Construct String from Binary Tree (建立一个二叉树的string)
You need to construct a string consists of parenthesis and integers from a binary tree with the preo ...
- [Swift]LeetCode606. 根据二叉树创建字符串 | Construct String from Binary Tree
You need to construct a string consists of parenthesis and integers from a binary tree with the preo ...
- 【LeetCode】606. Construct String from Binary Tree 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:先序遍历 日期 题目地址:https://l ...
随机推荐
- SpringMvc环境搭建(配置文件)
在上面的随笔里已经把搭建springmvc环境的基本需要的包都下下来了,拉下来就是写配置文件了. 下面左图是总的结构,右图是增加包 一.最开始当然是web.xml文件了,这是一个总的宏观配置 < ...
- Vue的组件
1,局部组件就是在Vue对象内部注册的构造器 <!DOCTYPE html> <html lang="en"> <head> <meta ...
- Sagit.Framework For IOS 开发框架入门教程6:网络请求STHttp
前言: IOS的文章,今天,再来补一篇,Sagit的教程: 虽然感觉IOS的文章没什么观众,还是努力写吧,-_-〜 Sagit 开源地址:https://github.com/cyq1162/Sagi ...
- Java 线程锁机制 -Synchronized Lock 互斥锁 读写锁
(1)synchronized 是互斥锁: (2)ReentrantLock 顾名思义 :可重入锁 (3)ReadWriteLock :读写锁 读写锁特点: a)多个读者可以同时进行读b)写者必须互斥 ...
- 云计算--网络原理与应用--20171122--STP与HSRP
简单了解STP 学习HSRP 实验 一. 简单学习STP STP(spanning tree protocol)生成树协议,就是把一个环形的结构改变成一个树形的结构.通过一些算法,在逻辑上阻塞一些端 ...
- 从0开始的LeetCode生活—461-Hamming Distance(汉明距离)
题目: The Hamming distance between two integers is the number of positions at which the corresponding ...
- class AClass<E extends Comparable>与class AClass<E extends Comaprable<E>>有什么区别?
new ArrayList<>()与new ArrayList()一样 都是为了做限定用的 如果不了解你可以看API 这个Comparable里面有一个方法compareTo(T o) 如 ...
- 标准C++类std::string的内存共享和Copy-On-Write(写时拷贝)
标准C++类std::string的内存共享,值得体会: 详见大牛:https://www.douban.com/group/topic/19621165/ 顾名思义,内存共享,就是两个乃至更多的对象 ...
- D的下L
D的小L 时间限制:4000 ms | 内存限制:65535 KB 难度:2 描述 一天TC的匡匡找ACM的小L玩三国杀,但是这会小L忙着哩,不想和匡匡玩但又怕匡匡生气,这时小L给 ...
- Node入门教程(3)第二章: Node 安装
Node 安装 官网下载地址: https://nodejs.org/en/download/ 安装方式 windows 下安装 建议直接选择:Windows Installer (.msi)下载进行 ...