1043. Is It a Binary Search Tree (25)

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Mq_Go

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分类专栏： PAT 文章标签： BST

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本文介绍了一个算法，用于判断给定的整数序列是否为二叉搜索树的预序遍历序列或是其镜像的预序遍历序列。通过构建二叉搜索树并比较给定序列，实现这一功能。

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1043. Is It a Binary Search Tree (25)

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

思路：

先按照他给出的顺序，构造一个二叉搜索树
然后分别后遍历，和前序
如果前序同题中给出相同
则为正确
输出后续


/*************************************************************
Author : qmeng
MailTo : qmeng1128@163.com
QQ     : 1163306125
Blog   : http://blog.youkuaiyun.com/Mq_Go/
Create : 2018-03-03 21:23:56
Version: 1.0
**************************************************************/
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
struct Tree{
    Tree *left;
    Tree *right;
    int data;
};

vector<int> pro,pos,proM,posM,origin;
void insert(Tree* root,int k){
    if(root == NULL){
        root = new Tree;
        root->data = k;
        root->left = NULL;
        root->right = NULL;
    }else{
        if(root->data > k )insert(root->left,k);
        else insert(root->right,k);
    }
    return;
}

void proTree(Tree *root){
    if(root==NULL)return;
    pro.push_back(root->data);
    proTree(root->left);
    proTree(root->right);
}
void posTree(Tree *root){
    if(root==NULL)return;
    pos.push_back(root->data);
    posTree(root->right);
    posTree(root->left);
}
void proMTree(Tree *root){
    if(root==NULL)return;
    proMTree(root->left);
    proMTree(root->right);
    proM.push_back(root->data);
}
void posMTree(Tree *root){
    if(root==NULL)return;
    posMTree(root->right);
    posMTree(root->left);
    posM.push_back(root->data);
} 
int main(){
    int n;
    cin >> n;
    Tree *root = NULL;
    for(int i = 0 ; i < n ; i++){
        int temp;
        cin >> temp;
        insert(root,temp);
        origin.push_back(temp);
    }
    proTree(root);
    posTree(root);
    proMTree(root);
    posMTree(root);

    if(pro == origin){
        printf("YES\n%d",proM[0]);
        for(int i = 1 ; i < n ; i++){
            printf(" %d",proM[i]);
        }
    }else if(pos == origin){
        printf("YES\n%d",posM[0]);
        for(int i = 1 ; i < n ; i++){
            printf(" %d",posM[i]);
        }
    }else{
        printf("NO\n");
    }
}