pat04-树7. Search in a Binary Search Tree (25)

04-树7. Search in a Binary Search Tree (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

To search a key in a binary search tree, we start from the root and move all the way down, choosing branches according to the comparison results of the keys. The searching path corresponds to a sequence of keys. For example, following {1, 4, 2, 3} we can find 3 from a binary search tree with 1 as its root. But {2, 4, 1, 3} is not such a path since 1 is in the right subtree of the root 2, which breaks the rule for a binary search tree. Now given a sequence of keys, you are supposed to tell whether or not it indeed correspnds to a searching path in a binary search tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N and M (<=100) which are the total number of sequences, and the size of each sequence, respectively. Then N lines follow, each gives a sequence of keys. It is assumed that the keys are numbered from 1 to M.

Output Specification:

For each sequence, print in a line "YES" if the sequence does correspnd to a searching path in a binary search tree, or "NO" if not.

Sample Input:

3 4
1 4 2 3
2 4 1 3
3 2 4 1

Sample Output:

YES
NO
NO

---

提交代码

 

法一：

如果是二叉查找树，树上除了叶结点，每个节点的左右子树必有一棵空，另一棵不空。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 using namespace std;
 8 struct node{
 9     int v;
10     node *l,*r;
11     node(){
12         l=r=NULL;
13     }
14 };
15 int main(){
16     //freopen("D:\\INPUT.txt","r",stdin);
17     int n,m;
18     scanf("%d %d",&n,&m);
19     while(n--){
20         queue<int> q;
21         int i,num;
22         node *p,*pp,*h=NULL;
23         for(i=0;i<m;i++){
24             scanf("%d",&num);
25             q.push(num);
26         }
27         h=new node();
28         h->v=q.front();
29         q.pop();
30         for(i=1;i<m;i++){
31             num=q.front();
32             q.pop();
33             p=h;
34             while(p){
35                 pp=p;//pp point to the father of p
36                 if(num>p->v&&p->l==NULL){
37                     p=p->r;
38                 }
39                 else{
40                     if(num<p->v&&p->r==NULL){
41                         p=p->l;
42                     }
43                     else
44                         break;//防止元素相等
45                 }
46             }
47             if(p==NULL){
48                 p=new node();
49                 p->v=num;
50                 if(num>pp->v){
51                     pp->r=p;
52                 }
53                 else{
54                     pp->l=p;
55                 }
56             }
57             else{
58                 break;
59             }
60         }
61         if(i==m){
62             printf("YES\n");
63         }
64         else{
65             printf("NO\n");
66         }
67     }
68     return 0;
69 }

 

 

法二：

先按题意建树，得到树后，用最后输入的数进行树的遍历，如果最后经过m次找到匹配的叶结点，说明序列正确；否则，序列错误。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 using namespace std;
 8 struct node{
 9     int v;
10     node *l,*r;
11     node(){
12         l=r=NULL;
13     }
14 };
15 int main(){
16     //freopen("D:\\INPUT.txt","r",stdin);
17     int n,m;
18     scanf("%d %d",&n,&m);
19     while(n--){
20         queue<int> q;
21         int i,num,wantfind;
22         node *p,*pp,*h;
23         for(i=0;i<m;i++){
24             scanf("%d",&num);
25             q.push(num);
26         }
27         wantfind=num;  //最后一个
28         h=new node();
29         h->v=q.front();
30         q.pop();
31         for(i=1;i<m;i++){ //建树
32             num=q.front();
33             q.pop();
34             p=h;
35             while(p){
36                 pp=p;//pp point to the father of p
37                 if(num>p->v){
38                     p=p->r;
39                 }
40                 else{
41                     p=p->l;
42                 }
43             }
44             p=new node();
45             p->v=num;
46             if(num>pp->v){
47                 pp->r=p;
48             }
49             else{
50                 pp->l=p;
51             }
52         }
53         int count=0;
54         p=h;
55         while(p){
56             if(wantfind>p->v){
57                 p=p->r;
58             }
59             else{
60                 p=p->l;
61             }
62             count++;
63         }
64         if(count==m){
65             printf("YES\n");
66         }
67         else{
68             printf("NO\n");
69         }
70     }
71     return 0;
72 }

 

转载于:https://www.cnblogs.com/Deribs4/p/4737182.html