题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3791
给一个序列，让你判断接下来输入的n个序列是否可以生成同一个二叉树。
BFS一遍就可以了。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 typedef struct Node {
23 Node* left;
24 Node* right;
25 int data;
26 Node() { left = NULL; right = NULL; }
27 }Node;
28
29 queue<Node*> q;
30
31 Node* insert(Node* cur, int data) {
32 if(cur == NULL) {
33 cur = new Node();
34 cur->data = data;
35 return cur;
36 }
37 if(data > cur->data) {
38 cur->right = insert(cur->right, data);
39 }
40 else {
41 cur->left = insert(cur->left, data);
42 }
43 return cur;
44 }
45
46 const int maxn = 11;
47 int n;
48 char tmp[maxn];
49 char a[maxn], b[maxn];
50
51 void bfs(char* x) {
52 Node* root = NULL;
53 scanf("%s", tmp);
54 int len = strlen(tmp);
55 for(int i = 0; i < len; i++) {
56 root = insert(root, tmp[i]-'0');
57 }
58 int cnt = 0;
59 while(!q.empty()) q.pop();
60 q.push(root);
61 while(!q.empty()) {
62 Node* tmp = q.front();
63 x[cnt++] = q.front()->data + '0';
64 if(tmp->left) q.push(tmp->left);
65 if(tmp->right) q.push(tmp->right);
66 q.pop();
67 }
68 }
69
70
71 int main() {
72 // freopen("in", "r", stdin);
73 while(~scanf("%d", &n) && n) {
74 memset(a, 0 , sizeof(a));
75 memset(b, 0 , sizeof(b));
76 bfs(a);
77 for(int i = 0; i < n; i++) {
78 bfs(b);
79 if(strcmp(a, b) == 0) {
80 printf("YES\n");
81 }
82 else {
83 printf("NO\n");
84 }
85 }
86 }
87 return 0;
88 }
来源：https://www.cnblogs.com/kirai/p/4908410.html