A1143 Lowest Common Ancestor (30 分| 二叉搜索树| 最近公共祖先，附详细注释，逻辑分析)-CFANZ编程社区

A1143 Lowest Common Ancestor (30 分| 二叉搜索树| 最近公共祖先，附详细注释，逻辑分析)

写在前面

思路分析

给出1棵2叉搜索树前序遍历，求结点u和v的最近共同祖先

map<int, bool> mp标记树中所有出现过的结点，遍历1遍pre数组，将当前结点标记为a

如果u和v分别在a的左、右，或者u、 v其中1个就是当前a，即(a >= u && a <= v) || (a >= v && a <= u)，
说明找到共同最低祖先a，退出当前循环

根据要求输出结果

学习ing

测试用例

input:
6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99

output:
LCA of 2 and 5 is 3.
8 is an ancestor of 7.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

ac代码

#include <iostream>
#include <vector>
#include <map>
using namespace std;
map<int, bool> mp;
int main()
{
    int m, n, u, v, a;
    scanf("%d %d", &m,&n);
    vector<int> pre(n);
    for(int i=0; i<n; i++)
    {
        scanf("%d", &pre[i]);
        mp[pre[i]] = true;
    }

    for(int i=0; i<m; i++)
    {
        scanf("%d %d", &u, &v);
        for(int j=0; j<n; j++)
        {
            a = pre[j];
            // BST特点，最近公共祖先
            if((a>=u && a<=v) || (a>=v && a<=u)) break;
        }

        if(mp[u] == false && mp[v] == false)
            printf("ERROR: %d and %d are not found.\n", u, v);
        else if (mp[u] == false || mp[v] == false)
            printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
        else if (a == u || a == v)
            printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
        else
            printf("LCA of %d and %d is %d.\n", u, v, a);
    }
    return 0;
}