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。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 using namespace std;
 struct node{
     int v;
     node *l,*r;
     node(){
         l=r=NULL;
     }
 };
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,m;
     scanf("%d %d",&n,&m);
     while(n--){
         queue<int> q;
         int i,num;
         node *p,*pp,*h=NULL;
         for(i=;i<m;i++){
             scanf("%d",&num);
             q.push(num);
         }
         h=new node();
         h->v=q.front();
         q.pop();
         for(i=;i<m;i++){
             num=q.front();
             q.pop();
             p=h;
             while(p){
                 pp=p;//pp point to the father of p
                 if(num>p->v&&p->l==NULL){
                     p=p->r;
                 }
                 else{
                     if(num<p->v&&p->r==NULL){
                         p=p->l;
                     }
                     else
                         break;//防止元素相等
                 }
             }
             if(p==NULL){
                 p=new node();
                 p->v=num;
                 if(num>pp->v){
                     pp->r=p;
                 }
                 else{
                     pp->l=p;
                 }
             }
             else{
                 break;
             }
         }
         if(i==m){
             printf("YES\n");
         }
         else{
             printf("NO\n");
         }
     }
     return ;
 }

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。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 using namespace std;
 struct node{
     int v;
     node *l,*r;
     node(){
         l=r=NULL;
     }
 };
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,m;
     scanf("%d %d",&n,&m);
     while(n--){
         queue<int> q;
         int i,num,wantfind;
         node *p,*pp,*h;
         for(i=;i<m;i++){
             scanf("%d",&num);
             q.push(num);
         }
         wantfind=num;  //最后一个
         h=new node();
         h->v=q.front();
         q.pop();
         for(i=;i<m;i++){ //建树
             num=q.front();
             q.pop();
             p=h;
             while(p){
                 pp=p;//pp point to the father of p
                 if(num>p->v){
                     p=p->r;
                 }
                 else{
                     p=p->l;
                 }
             }
             p=new node();
             p->v=num;
             if(num>pp->v){
                 pp->r=p;
             }
             else{
                 pp->l=p;
             }
         }
         int count=;
         p=h;
         while(p){
             if(wantfind>p->v){
                 p=p->r;
             }
             else{
                 p=p->l;
             }
             count++;
         }
         if(count==m){
             printf("YES\n");
         }
         else{
             printf("NO\n");
         }
     }
     return ;
 }