1043 Is It a Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/daybreak_alonely/article/details/127806552

这篇博客讲解如何通过算法判断一组整数序列是否为二叉搜索树的前序遍历或其镜像翻转结果，并提供C++代码实现。涉及分类讨论、递归遍历和镜像操作。

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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

题目大意

给你一串数组，判断这是否是对⼀棵⼆叉搜索树或其镜像翻转后，进⾏前序遍历的结果

是的话输出 YES + 该树的后续遍历

否则 NO

思路

分类讨论是否为镜像翻转

C/C++

#include<bits/stdc++.h>
using namespace std;
vector<int> tree1,tree2;
int N,num;
bool flag = true; // 判断有无镜像翻转
void create(int head,int tail);
int main()
{
    cin >> N;
    for(int z=0;z<N;z++){
        cin >> num;
        tree1.push_back(num);
    }
    create(0,N-1);
    if(tree2.size()!=tree1.size()){
        tree2.clear();
        flag = false;
        create(0,N-1);
    }

    if(tree2.size()!=tree1.size()) puts("NO");
    else
    {
        cout << "YES" << endl << tree2[0];
        for(int z=1;z<N;z++) cout << " " << tree2[z];
    }
    return 0;
}
void create(int head,int tail)
{
    if(head>tail) return;
    int left=head+1,right=tail;
    if(flag){
        while (left<=tail && tree1[head]>tree1[left]) left++;
        while (right>head && tree1[head]<=tree1[right]) right--;
    }else{
        while (left<=tail && tree1[head]<=tree1[left]) left++;
        while (right>head && tree1[head]>tree1[right]) right--;
    }
    if(left-right>1) return;
    create(head+1,left-1);
    create(right+1,tail);
    tree2.push_back(tree1[head]);
}