1043. Is It a Binary Search Tree (25)

1043. Is It a Binary Search Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line "YES" if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or "NO" if not. Then if the answer is "YES", print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

7 8 6 5 7 10 8 11

Sample Output 1:

YES 5 7 6 8 11 10 8

Sample Input 2:

7 8 10 11 8 6 7 5

Sample Output 2:

YES 11 8 10 7 5 6 8

Sample Input 3:

7 8 6 8 5 10 9 11

Sample Output 3:

NO

这道题在自己做的时候，想避免构建树，利用递归的算法做，但最后有几个检测点过不了。网上看到一个比较好的做法https://blog.youkuaiyun.com/IAccepted/article/details/20539395，先根据序列创建BST，再先序遍历和镜像的先序遍历（原树的根右左遍历），将序列与得到的两个遍历序列比较。

#include <cstdio>  
#include <cstdlib>  
#include<vector> 
using namespace std;  
  
const int maxx = 1002;  
  
typedef struct Tree{  
    Tree * left;  
    Tree * right;  
    int ele;  
}Tree;  
  
vector<int> righta,rightb;  
vector<int> post,seq;  
  
void insert(Tree *&root,int num){  
    if(root==NULL){  
        root = (Tree *)malloc(sizeof(Tree));  
        root->ele = num;  
        root->left = NULL;  
        root->right = NULL;  
        return;  
    }  
  
    if(num<root->ele){  
        insert(root->left,num);  
    }else{  
        insert(root->right,num);  
    }  
}  
  
void preOrder(Tree *root){//前序  
    if(root==NULL)return;  
    righta.push_back(root->ele);  
    preOrder(root->left);  
    preOrder(root->right);  
}  
  
void ARL(Tree *root){//根右左mirror的前序  
    if(root==NULL)return;  
    rightb.push_back(root->ele);  
    ARL(root->right);  
    ARL(root->left);  
}  
  
void postOrder(Tree *root){//后序  
    if(root==NULL)return;  
    postOrder(root->left);  
    postOrder(root->right);  
    post.push_back(root->ele);  
}  
  
void print(vector<int> vc){  
    vector<int>::iterator ite = vc.begin();  
  
    bool flg = true;  
    for(;ite!=vc.end();++ite){  
        if(!flg){  
            printf(" %d",*ite);  
        }else{  
            printf("%d",(*ite));  
            flg = false;  
        }  
    }  
    printf("\n");  
}  
  
void change(Tree *root){  
    if(root==NULL)return;  
    Tree *temp = root->left;  
    root->left = root->right;  
    root->right = temp;  
    change(root->left);  
    change(root->right);  
}  
  
int main(){  
    int n,i,num;  
    Tree *root = NULL;  
    scanf("%d",&n);  
    for(i=0;i<n;++i){  
        scanf("%d",&num);  
        seq.push_back(num);  
        insert(root,num);  
    }  
      
    preOrder(root);  
    ARL(root);  
    if(righta==seq){  
        printf("YES\n");  
        post.clear();  
        postOrder(root);  
        print(post);  
    }else if(rightb==seq){  
        printf("YES\n");  
        change(root);  
        post.clear();  
        postOrder(root);  
        print(post);  
    }else{  
        printf("NO\n");  
    }  
    return 0;  
}