题目来源：https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312
备考汇总贴：2020年3月PAT甲级满分必备刷题技巧
题目
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 (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), 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.题目大意（以下内容主要转载自liuchuo，补充了讲解）
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
题目分析
1.二叉搜索树的前序遍历有个特点：排在前面的某个节点一定是“在它后面且不大于它的节点”的父亲。所以，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
2.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
满分代码
#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];
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;
}其他解法（转载自其他网友）
建树、递归查找节点看看符不符合lca的定义。
#include<iostream>
#include<unordered_map>
#include<algorithm>
#include<unordered_set>
#include<string>
using namespace std;
#define MAX 10005
int pre[MAX],in[MAX],N,M;
typedef struct node{
int left,right;
}node;
unordered_map<int,node>arr;
void CreatBST(int preL,int inL,int inR)
{
if(inL>inR)return ;
int i=inL;
while(pre[preL]!=in[i])++i;
if(i-inL)arr[pre[preL]].left=pre[preL+1];
else arr[pre[preL]].left=0;
if(inR-i)arr[pre[preL]].right=pre[preL+i-inL+1];
else arr[pre[preL]].right=0;
CreatBST(preL+1,inL,i-1);
CreatBST(preL+i-inL+1,i+1,inR);
}
void findLCA(int v,int a,int b)
{
if(a==v)printf("%d is an ancestor of %d.\n",a,b);
else if(b==v)printf("%d is an ancestor of %d.\n",b,a);
else if(a<v&&b>v||a>v&&b<v)printf("LCA of %d and %d is %d.\n",a,b,v);
else if(a<v&&b<v)findLCA(arr[v].left,a,b);
else findLCA(arr[v].right,a,b);
}
int main()
{
//freopen("test.txt","r",stdin);
int i,u,v,root;
scanf("%d %d",&M,&N);
for(i=0;i<N;++i){
scanf("%d",&pre[i]);
if(i==0)root=pre[0];
in[i]=pre[i];
}
sort(in,in+N);
CreatBST(0,0,N-1);
while(M--){
scanf("%d %d",&u,&v);
if(!arr.count(u)&&!arr.count(v))printf("ERROR: %d and %d are not found.\n",u,v);
else if(!arr.count(u)||!arr.count(v))printf("ERROR: %d is not found.\n",arr.count(u)?v:u);
else findLCA(root,u,v);
}
return 0;
}