标签：ogr typedef binary lib ram pairs tput algorithm amp
Trees are fundamental in many branches of computer science (Pun definitely intended). Current stateof-the
art parallel computers such as Thinking Machines’ CM-5 are based on fat trees. Quad- and
octal-trees are fundamental to many algorithms in computer graphics.
This problem involves building and traversing binary trees.
Given a sequence of binary trees, you are to write a program
that prints a level-order traversal of each tree. In this
problem each node of a binary tree contains a positive integer
and all binary trees have have fewer than 256 nodes.
In a level-order traversal of a tree, the data in all nodes at
a given level are printed in left-to-right order and all nodes at
level k are printed before all nodes at level k + 1.
For example, a level order traversal of the tree on the right
is: 5, 4, 8, 11, 13, 4, 7, 2, 1.
In this problem a binary tree is specified by a sequence
of pairs ‘(n,s)’ where n is the value at the node whose path
from the root is given by the string s. A path is given be
a sequence of ‘L’s and ‘R’s where ‘L’ indicates a left branch and ‘R’ indicates a right branch. In the
tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is
specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from
the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node
paths in the tree is given a value exactly once.
Input
The input is a sequence of binary trees specified as described above. Each tree in a sequence consists
of several pairs ‘(n,s)’ as described above separated by whitespace. The last entry in each tree is ‘()’.
No whitespace appears between left and right parentheses.
All nodes contain a positive integer. Every tree in the input will consist of at least one node and
no more than 256 nodes. Input is terminated by end-of-file.
Output
For each completely specified binary tree in the input file, the level order traversal of that tree should
be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a
node is given a value more than once, then the string ‘not complete’ should be printed.
Sample Input
(11,LL) (7,LLL) (8,R)
(5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
(3,L) (4,R) ()
Sample Output
5 4 8 11 13 4 7 2 1
not complete
题目链接：https://vjudge.net/problem/UVA-122
(11,LL)表示节点值为11，在第二层的左，第三层的左；
(5,)表示根节点;
总之，每一个L或者R代表从root到达该节点在每一层的转向,问，给一串这样的字符串，输出二叉树层序遍历的结果，
如果输入重复或者节点的val为0则表示输出错误“not complete”.
根据输入构造二叉树，然后用queue层序遍历输出即可。
（1）根据输入创建二叉树，技巧：sscanf(&s[1], "%d", &v),从数组中读取数字，strchr(s, ‘,‘)+1获取数字后的字符串最后一个‘)‘在处理时忽略;
（2）指针实现时，记得deletenode,root要先newnode;
（3）数组实现时，各节点按顺序存放，通过lf[maxn]和rt[maxn]数组确定左右节点；
（4）注意，已经有值的节点重复赋值要另flag = false,以及层序遍历时出现未赋值节点也要flag = false.
分析：1。在输入方面，直接开了一个bool read_input（）函数，这样在输入EOF时，就返回false，所以可以在main函数里直接这样写：while(read_input()){...}。
2。在这个read_input 函数里面，用到了sscanf （&s[1]），这样就把字符串从位置1开始的字符串后面读取整数！，后面又用到了strchr（s,‘,‘）把逗号后面的字符串提取出来。
3。在结构体node里面，用了一个构造函数，不仅把have_value这个关键变量初始化为false，并且把一个node 里面的left,right也初始化到NULL，在前面学字典树时，就没这样做，而是new node后再用循环初始化NULL，这里直接用构造函数初始化，就感觉比较巧妙了。
3。遍历二叉树时用了bfs+队列，先判断left，在判断right，这样就保证了输出顺序的正确性，
4。最后释放内存用到了递归释放内存，类似于字典树。
： //#include <bits/stdc++.h> #include<stdio.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<string.h> #include<string> #include<stack> #include<queue> #include<iostream> #define INF 0xffffffff using namespace std; typedef long long ll; const int maxn = 300; char s[maxn]; bool failed; vector<int> a; struct Node{ bool have_value; int v; Node *left, *right; Node():have_value(false), left(NULL), right(NULL){} }; Node* root; Node* newnode(){ return new Node(); } void addnode(int v, char* s){ int n = strlen(s); Node *u = root; for(int i=0; i<n; i++){ if(s[i] == ‘L‘){ if(u->left == NULL) u->left = newnode(); u = u->left; } else if(s[i] == ‘R‘){ if(u->right == NULL) u->right = newnode(); u = u->right; } } if(u->have_value) failed = true; u->v = v; u->have_value = true; } void remove_tree(Node* u){ if(u == NULL) return; remove_tree(u->left); remove_tree(u->right); delete u; } bool read_input(){ failed = false; remove_tree(root); root = newnode(); for(;;){ if(scanf("%s", s) != 1) return false; if(!strcmp(s, "()")) break; int v; sscanf(&s[1], "%d", &v); addnode(v, strchr(s, ‘,‘)+1); } return true; } bool bfs(vector<int>& ans){ queue<Node*> q; ans.clear(); q.push(root); while(!q.empty()){ Node *u = q.front(); q.pop(); if(!u->have_value) return false; ans.push_back(u->v); if(u->left != NULL) q.push(u->left); if(u->right != NULL) q.push(u->right); } return true; } int main(){ while(read_input()){ if(bfs(a) && !failed){ for(int i=0; i<a.size(); i++){ if(i >0) printf(" "); printf("%d", a[i]); } printf("\n"); } else{ printf("not complete\n"); } } return 0; }
数组实现代码
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<string.h> #include<string> #include<stack> #include<queue> #include<iostream> using namespace std; typedef long long ll; const int maxn=300; const int root=1; int lf[maxn]; int rt[maxn]; bool hv[maxn]; int val[maxn]; vector<int> ans; bool flag; int cnt; void newtree() { cnt=root; lf[root]=rt[root]=0; hv[root]=false; } int newnode() { int u=++cnt; lf[u]=rt[u]=0; hv[u]=false; return u; } void addnode(int v,char *s) { int n=strlen(s); int u=root; for(int i=0; i<n; i++) { if(s[i]==‘L‘) { if(lf[u]==0) lf[u]=newnode();//节点不存在，建立新节点 u=lf[u];//往左走 } else if(s[i]==‘R‘) { if(rt[u]==0) rt[u]=newnode(); u=rt[u]; }//忽略其他情况 即最后那个多余的右括号 } if(hv[u]) flag=false;//have_value 本身已经有值 val[u]=v; hv[u]=true; } bool input() { newtree(); flag=true; char s[30]; for(;;) { if(scanf("%s",s)!=1) return false;//输入格式不正确 if(!strcmp(s,"()")) break;//输入结束 int v; //（11,LL) s[1]指 11,LL) sscanf(&s[1],"%d",&v);//读入节点值 //保存后面的 (11,LL) 保存s[1]保存LL) //strchr返回的是从左往右 第一个字符”，“的指针 addnode(v,strchr(s,‘,‘)+1);//添加节点 } return true; } bool bfs() //宽度优先遍历 { ans.clear(); queue<int> q; q.push(root); while(!q.empty()) { int u=q.front(); q.pop(); if(hv[u]==false) return false; ans.push_back(val[u]); if(lf[u]!=0) q.push(lf[u]); if(rt[u]!=0) q.push(rt[u]); } return true; } int main() { while(input()){ if(bfs()&&flag){ for(int i=0;i<ans.size();i++) { if(i>0) printf(" "); printf("%d",ans[i]); } printf("\n"); } else printf("not complete\n"); } }
本题是训练二叉树的一道好题。首先要解决读数据问题，根据题意，当输入为“()”时，结束该组数据读入，当没有字符串时，整个输入结束。因此可以专门编写一个readin()函数，类型设置为bool型，遇到第一种情况时返回true，遇到第二种情况返回false，主程序中只要发现readin返回false时就break，结束整个大循环。
接下来要建立二叉树，首先为二叉树编写一个结构体，然后根据字符串的输入情况来建树，如果是‘L’就往左走，走不动时建一颗新树，同样的方法处理右子树，最后读入结点值。由于输入可能有误，因此用一个全局变量failed来记录是否有输入错误的情况出现，如果在建树过程中发现该结点已经被赋过值，那么全局变量failed变为true。
最后开始BFS找结点值，此时可能出现有结点没有结点值的情况，因此要把bfs定义为bool型，只要出现这种非法情况，返回false。最后便不难根据情况进行输出了。
标签：ogr typedef binary lib ram pairs tput algorithm amp
原文地址：http://www.cnblogs.com/is-Tina/p/7417689.html
