【ACM】- PAT. A1043 Is It a Binary Search Tree 【BST】

题目链接
题目分析

给出一个序列,判断是否为BST树的先序遍历序列

解题思路

1、按照所给序列构造BST
2、先序遍历这棵树,得到的序列与所给序列相同则说明是,否则不是
注:用vector保存序列,这样比较时比较方便,不然还要写循环;


AC程序(C++)
/**********************************
*@ID: 3stone
*@ACM: PAT.A1043 Is It a Binary Search Tree
*@Time: 18/8/13
*@IDE: VSCode 2018 + clang++
***********************************/
#include<cstdio>
#include<vector>

using namespace std;

//结点
struct node{
    int data;
    node *lchild, *rchild;
};

int N;
vector<int> org, pre, pre_mirror, post, post_mirror;

node* new_node(int x) {
    node* root = new node;
    root->data = x;
    root->lchild = root->rchild = NULL;
    return root;
}

void insert(node* &root, int data_x) {
    if(root == NULL){
        root = new_node(data_x);
        return;
    }
    if(data_x < root->data) {
        insert(root->lchild, data_x);
    } else {
        insert(root->rchild, data_x);
    }
}

void pre_order_normal(node *root) {
    if(root == NULL) return;
    pre.push_back(root->data);
    pre_order_normal(root->lchild);
    pre_order_normal(root->rchild);
}

void pre_order_mirror(node *root) {
    if(root == NULL) return;
    pre_mirror.push_back(root->data);
    pre_order_mirror(root->rchild);
    pre_order_mirror(root->lchild);
}

void post_order_normal(node* root) {
    if(root == NULL) return;
    post_order_normal(root->lchild);
    post_order_normal(root->rchild);
    post.push_back(root->data);
}

void post_order_mirror(node* root) {
    if(root == NULL) return;
    post_order_mirror(root->rchild);
    post_order_mirror(root->lchild);
    post_mirror.push_back(root->data);
}

//输出保存在vector中的序列
void output_seq(vector<int> vec) {
    int num = 0;
    for(int i = 0 ; i < vec.size(); i++) {
        if(++num == N)
            printf("%d\n", vec[i]);
        else
            printf("%d ", vec[i]);
    }
}

int main() {
    int key;
    while(scanf("%d", &N) != EOF) {
        node* root = NULL;
        org.clear();
        pre.clear();
        pre_mirror.clear();
        post.clear();
        post_mirror.clear();
        for(int i = 0; i < N; i++) {
            scanf("%d", &key);
            insert(root, key);
            org.push_back(key);
        }

        pre_order_normal(root); 

        if(pre == org) {

            printf("YES\n");
            post_order_normal(root);
            output_seq(post);

        } else {

            pre_order_mirror(root);
            if(pre_mirror == org) {
                printf("YES\n");
                post_order_mirror(root); 
                output_seq(post_mirror);
            } else {
                printf("NO\n");
            }

        } //else 

    }//while

    return 0;
}

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转载自blog.csdn.net/qq_26398495/article/details/81772890
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