SPOJ NICEBTRE - Nice Binary Trees(树 先序遍历)
Description
Binary trees can sometimes be very difficult to work with. Fortunately, there is a class of trees with some really nice properties. A rooted binary tree is called “nice”, if every node is either a leaf, or has exactly two children.
For example, the following tree is nice,
but the following tree is not.
The leaves of a nice binary tree are labeled by the letter ‘l’, and other nodes are labeled by the letter ‘n’.
Given the pre-order traversal of a nice binary tree, you are required to find the depth of the tree.
Notes :
1. The depth of a tree is defined as the length of the longest path with one end at the root.
2. The pre-order traversal of the tree in the first image above produces the string “nlnnlll”.
Input
The first line contains the number of test cases T. T lines follow. Each line contains a string, which represents the pre-order traversal of a “nice” binary tree. Leaves are represented by the letter ‘l’ and other nodes by the letter ‘n’. The input is guaranteed to be the preorder traversal of a nice binary tree.
0 < T < 20
Length of the input string in each test case is at most 10000.
Output
Output one line for each test case, containing a single integer, the depth of tree.
Sample Input
3
l
nlnll
nlnnlll
Sample Output
0
2
3
#include<bits/stdc++.h>
using namespace std;
const int maxn = 10005;
char str[maxn];
int pos,height; void dfs(int depth){
pos++;
height = max(depth,height);
if (str[pos] == '\0') return;
if (str[pos] == 'l') return;
if (str[pos] == 'n'){
dfs(depth + 1); //left
dfs(depth + 1); //right;
}
} int main(){
int T;
scanf("%d",&T);
while (T--){
memset(str,0,sizeof(str));
height = 0,pos = -1;
scanf("%s",str);
dfs(0);
cout << height << endl;
}
return 0;
}
SPOJ NICEBTRE - Nice Binary Trees(树 先序遍历)的更多相关文章
- [LeetCode] 538. 把二叉搜索树转换为累加树 ☆(中序遍历变形)
把二叉搜索树转换为累加树 描述 给定一个二叉搜索树(Binary Search Tree),把它转换成为累加树(Greater Tree),使得每个节点的值是原来的节点值加上所有大于它的节点值之和. ...
- [jzoj 5662] 尺树寸泓 解题报告 (线段树+中序遍历)
interlinkage: https://jzoj.net/senior/#contest/show/2703/1 description: solution: 发现$dfs$序不好维护 注意到这是 ...
- Bzoj 3173: [Tjoi2013]最长上升子序列 平衡树,Treap,二分,树的序遍历
3173: [Tjoi2013]最长上升子序列 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 1183 Solved: 610[Submit][St ...
- PAT甲级|1151 LCA in a Binary Tree 先序中序遍历建树 lca
给定先序中序遍历的序列,可以确定一颗唯一的树 先序遍历第一个遍历到的是根,中序遍历确定左右子树 查结点a和结点b的最近公共祖先,简单lca思路: 1.如果a和b分别在当前根的左右子树,当前的根就是最近 ...
- Tree(树的还原以及树的dfs遍历)
紫书:P155 uva 548 You are to determine the value of the leaf node in a given binary tree that is th ...
- 由后序遍历结果构造二叉查找树 && 二叉查找树链表化
二叉查找树通俗说就是左孩子比父亲小,右孩子比父亲大.构造这么一个树,树嘛,递归即可. 例如一棵树后序遍历是这样(下图的树):2 9 8 16 15 10 25 38 45 42 30 20.最后的20 ...
- PAT L2-004. 这是二叉搜索树吗?【前序遍历转化为后序遍历】
一棵二叉搜索树可被递归地定义为具有下列性质的二叉树:对于任一结点, 其左子树中所有结点的键值小于该结点的键值: 其右子树中所有结点的键值大于等于该结点的键值: 其左右子树都是二叉搜索树. 所谓二叉搜索 ...
- 【树】Flatten Binary Tree to Linked List(先序遍历)
题目: Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 ...
- PAT树_层序遍历叶节点、中序建树后序输出、AVL树的根、二叉树路径存在性判定、奇妙的完全二叉搜索树、最小堆路径、文件路由
03-树1. List Leaves (25) Given a tree, you are supposed to list all the leaves in the order of top do ...
随机推荐
- CSS3进度条动画
<!DOCTYPE html> <html> <head> <meta charset="utf-8" /> <title&g ...
- Viewer.js – 强大的JS/jQuery图片查看器
简介 Viewer.js 是一款强大的图片查看器,像门户网站一般都会有各自的图片查看器,如果您正需要一款强大的图片查看器,也许 Viewer.js 是一个很好的选择.Viewer.js 有以下特点: ...
- Python的五大数据类型的作用、定义方式、使用方法
一.简述Python的五大数据类型的作用.定义方式.使用方法: 1. 数字类型int: 1.整形 作用:可以表示人的年龄,身份证号码,身高和体重等 定义方式: weight = 130 print( ...
- Node 12 值得关注的新特性
前言 时隔一年,Node.js 12 如约而至,正式发布第一个 Current 版本. 该版本带来了诸如: V8 更新带来好多不错的特性. HTTP 解析速度提升. 启动速度大幅提升. 更好的诊断报告 ...
- Docker实战部署应用——Tomcat
Tomcat 部署 拉取tomcat镜像 docker pull tomcat:8 创建tomcat容器 创建tomcat容器用于 Web应用,并且进行目录映射 docker run -id --na ...
- 【vue】父子组件间通信----传函数
(一)子组件 调用 父组件 方法 方式一) 子组件中通过this.$parent.event来调用父组件的方法 父组件 <template> <div> <child&g ...
- idea import class的快捷键
有自动import class的快捷键 设置如下: 1.alt+enter 2.写好代码之后ctrl+alt+l格式化代码,优化导入包 1.alt+enter 2.写好代码之后ctrl+alt+l ...
- (转)nginx+redis实现接入层高性能缓存技术
转自:https://blog.csdn.net/phil_code/article/details/79154271 1. OpenRestyOpenResty是一个基于 Nginx与 Lua的高性 ...
- hdu 4992 Primitive Roots 【求原根模板】
题目链接 大题流程: 判定是否有原根->求出最小原根->利用最小原根找出全部原根 #include<bits/stdc++.h> using namespace std; ty ...
- 【leetcode】1054. Distant Barcodes
题目如下: In a warehouse, there is a row of barcodes, where the i-th barcode is barcodes[i]. Rearrange t ...