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=0;i<m;i++){
            scanf("%d",&num);
            q.push(num);
        }
        h=new node();
        h->v=q.front();
        q.pop();
        for(i=1;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 0;
}

 

 

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=0;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=1;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=0;
        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 0;
}