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
题目类型：二叉树遍历
算法分析：首先应该先建树，然后按照题目的要求边读入数据，边向二叉树添加节点。对于没有的子节点则创建一个节点，有子节点的则直接移动到其子节点中，直到输入命令结束。判断所在的节点是否已经赋过值了，如果已经赋过值了就置合法标志valid为false。接着使用BFS遍历建好的二叉树并将节点的值赋值到输出数组中，对于没有赋过值的节点，则将valid标志置为false并break
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/************************************************** filename :p.cpp author :maksyuki created time :2018/1/29 8:59:35 last modified :2018/1/29 9:29:35 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 666; int Left[maxn], Right[maxn], val[maxn], rt, cnt; bool have_val[maxn], fail; void BuildTree() { rt = 1; memset(Left, -1, sizeof(Left)); memset(Right, -1, sizeof(Right)); memset(have_val, false, sizeof(have_val)); cnt = 2; } stringstream ss; void AddNode(const string& sa, const string& vb) { ss.clear(); ss << sa; int v; ss >> v; int p = rt; for(int i = 0; vb[i]; i++) { if(vb[i] == 'L') { if(Left[p] == -1) Left[p] = cnt++; p = Left[p]; } else if(vb[i] == 'R') { if(Right[p] == -1) Right[p] = cnt++; p = Right[p]; } } if(have_val[p]) { fail = true; return; } val[p] = v; have_val[p] = true; } vector<int> ans; void Output() { ans.clear(); queue<int> qu; qu.push(rt); while(!qu.empty() && !fail) { int tmp = qu.front(); qu.pop(); if(!have_val[tmp]) { fail = true; break; } ans.emplace_back(val[tmp]); if(Left[tmp] != -1) qu.push(Left[tmp]); if(Right[tmp] != -1) qu.push(Right[tmp]); } if(!fail) { int alen = ans.size(); for(int i = 0; i < alen; i++) { if(!i) cout << ans[i]; else cout << " " << ans[i]; } cout << endl; } else cout << "not complete" << endl; } int main() { #ifdef LOCAL CFF; //CFO; #endif string s, sa, sb; while(cin >> s) { BuildTree(); fail = false; do { if(s == "()") break; sa = s.substr(1, s.find(',') - 1); sb = s.substr(s.find(',') + 1); AddNode(sa, sb); }while(cin >> s); Output(); } return 0; } |
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