1143 Lowest Common Ancestor (30 分)（二叉查找树）

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.
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.
Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.
Output Specification:
For each given pair of U and V, print in a line “LCA of U and V is A.” if the LCA is found and A is the key. But if A is one of U and V, print “X is an ancestor of Y.” where X is A and Y is the other node. If U or V is not found in the BST, print in a line “ERROR: U is not found.” or “ERROR: V is not found.” or “ERROR: U and V are not found.”.

Sample Input:

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

Sample 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.

生词

英文	解释
descendants	后代

题目大意：

给出一棵二叉搜索树的前序遍历，问结点u和v的共同最低祖先是谁～

分析：

map<int, bool> mp用来标记树中所有出现过的结点，遍历一遍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其中一个就是当前a，说明找到了这个共同最低祖先a，退出当前循环～最后根据要求输出结果即可～

原文链接：https://blog.csdn.net/liuchuo/article/details/79618849

题解

根据该博主的分析，已知BST前序构建BST

心急吃不了热豆腐，慢慢来吧

#include <bits/stdc++.h>

using namespace std;
map<int,bool> mp;
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int m,n,u,v,a;
    cin>>m>>n;
    vector<int> pre(n);
    for(int i=0;i<n;i++){
        cin>>pre[i];
        mp[pre[i]]=true;
    }
    for(int i=0;i<m;i++){
        cin>>u>>v;
        for(int j=0;j<n;j++){
            a=pre[j];
            if((a>u&&a<v)||(a>v&&a<u)||(a==u)||(a==v)) 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;
}