Trees on the level
Background
Trees are fundamental in many branches of computer science. Current state-of-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.
The Problem
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 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
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.
The 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 roght 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 treminated by end-of-file.
The 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
-----------------------------------------------------------------------------------------------------
思路解析：
1.数据结构的选取
本题的数据元素为若干个结点，选取数组在本题中是行不通的，因为考虑到极限情况，256个结点形成一条链，那样的开销浪费将会是巨大的，数组也开不下。看来，需要采用动态结构，根据需要建立新的结点，然后将其组织成一棵完整的树。
2.功能模块划分（每个模块用一个函数来实现，函数名需要起的有意义）
本题要求实现以下几个功能：①数据输入②建树③层次遍历树并输出对应结点值
-----------------------------------------------------------------------------------------------------
代码1：（有BUG，你能知道错在哪吗?）
1 #include <cstdio>
2 #include <cstring>
3 #include <vector>
4 #include <queue>
5 using namespace std;
6 const int maxn = 300;
7 //结点类型
8 struct Node
9 {
10 bool have_value; //是否被赋过值
11 int v; //结点值
12 Node *left, *right;
13 Node():have_value(false),left(NULL),right(NULL){} //构造函数
14 };
15
16 Node* root = NULL; //二叉树的根结点
17 char s[maxn]; //保存读入结点
18 bool failed;
19 Node* newnode()
20 {
21 return new Node();
22 }
23
24 void addnode(int v, char* s)
25 {
26 // printf("%s \n", s);
27 int n = strlen(s);
28 // printf("%d \n", strlen(s));
29 Node* u = root; //从根结点开始往下走
30 for (int i = 0; i < n; i++)
31 {
32 if (s[i] == ‘L‘)
33 {
34 if (u->left == NULL)
35 {
36 u->left = newnode(); //结点不存在，建立新结点
37 u = u->left; //往左走
38 }
39
40 }
41 else if (s[i] == ‘R‘)
42 {
43 if (u->right == NULL)
44 {
45 u->right = newnode();
46 u = u->right;
47 }
48
49 }
50
51
52 }
53 #if 1
54 if (u->have_value) //已经赋过值，表明输入有误
55 {
56 failed = true;
57 }
58 #endif
59 u->v = v;
60 u->have_value = true; //别忘记做标记
61 }
62
63 void remove_tree(Node* u)
64 {
65 if (u == NULL)
66 {
67 return; //提前判断比较稳妥
68 }
69 remove_tree(u->left); //递归释放左子树
70 remove_tree(u->right); //递归释放右子树
71 delete u; //调用u的析构函数并释放u结点本身的内存
72 }
73
74 bool read_input()
75 {
76 // int count = 1;
77 failed = false;
78 remove_tree(root); //释放上一次构建的二叉树
79 root = newnode(); //创建根结点
80 for (;;)
81 {
82 if (scanf("%s", s) != 1)
83 {
84 return false; //整个输入结束
85 }
86 if (!strcmp(s, "()")) //读到结束标志，退出循环
87 {
88 break;
89 }
90 // printf("%s \n", s);
91 int v;
92 sscanf(&s[1], "%d", &v); //读入结点值
93 // printf("%d : v = %d\n", count++, v);
94 addnode(v, strchr(s, ‘,‘)+1);
95 //printf("%s \n", strchr(s, ‘,‘)+1);
96 }
97 // printf("root value = %d \n", root->v);
98 return true;
99 }
100
101 bool bfs(vector<int>& ans)
102 {
103 queue<Node*> q;
104 ans.clear();
105 q.push(root); //初始时只有一个根结点
106 while (!q.empty())
107 {
108 Node* u = q.front();
109 q.pop();
110 if (!u->have_value)
111 {
112 return false; //有结点没有被赋过值，表明输入有误
113 }
114 ans.push_back(u->v);//增加到输出序列尾部
115 //printf("%d \n", u->v);
116 if (u->left != NULL)
117 {
118 q.push(u->left); //把左结点（如果有）放入队列
119 }
120 if (u->right != NULL)
121 {
122 q.push(u->right); //把右结点（如果有）放入队列
123 }
124 }
125 return true; //输入正确
126 }
127 /*
128 int main()
129 {
130 while (1)
131 {
132 if (read_input())
133 {
134 vector<int> ans;
135 bfs(ans);
136 for (int i = 0; i < ans.size(); i++)
137 {
138 printf("%d ", ans[i]);
139 }
140 printf("\n");
141 }
142 else
143 {
144 printf("%d\n", -1);
145 }
146 #if 0
147 printf("ans.size() == %d\n", ans.size());
148 #endif
149
150
151 return 0;
152 }
153 */
154 int main()
155 {
156 vector<int> ans;
157 while (read_input())
158 {
159 if (!bfs(ans))
160 {
161 failed = true;
162 }
163 if (failed)
164 {
165 printf("not complete\n");
166 }
167 else
168 {
169 for (int i = 0; i < ans.size(); i++)
170 {
171 if (i != 0)
172 {
173 printf(" ");
174 }
175 printf("%d", ans[i]);
176 }
177 printf("\n");
178 }
179 }
180
181 return 0;
182 }
代码2：（正确代码）
链接：https://github.com/ToWorld/OJ/blob/master/UVa122