大致题意就是给出一棵树的先序、中序遍历序列，可以构造一棵树。然后给两个顶点，找出这两个结点的最近公共祖先。

 1 #include<iostream>
 2 #include<unordered_map>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn = 10010;
 6 int m,n,pre[maxn],in[maxn],u,v,flag;
 7 unordered_map<int,int> pos;
 8 
 9 void LCA(int inL,int inR,int preRoot,int u,int v) {
10     if(inL > inR) return ;
11     int inRoot = pos[pre[preRoot]],inU = pos[u],inV = pos[v];
12     if((inU < inRoot && inRoot < inV)||(inU > inRoot && inRoot > inV))
13         printf("LCA of %d and %d is %d.\n",u,v,in[inRoot]);
14     else if(inU == inRoot)
15         printf("%d is an ancestor of %d.\n",u,v);
16     else if(inV == inRoot)
17         printf("%d is an ancestor of %d.\n",v,u);
18     else if(inU < inRoot && inV < inRoot) //遍历左子树 
19         LCA(inL,inRoot-1,preRoot+1,u,v);
20     else if(inU > inRoot && inV > inRoot) //遍历右子树 
21         LCA(inRoot+1,inR,preRoot+1+(inRoot-inL),u,v);
22 }
23 
24 int main() {
25     cin>>m>>n;
26     for(int i = 1; i <= n; ++i) scanf("%d",&in[i]),pos[in[i]] = i;
27     for(int i = 1; i <= n; ++i) scanf("%d",&pre[i]);
28     for(int i = 0; i < m; ++i) {
29         scanf("%d%d",&u,&v);
30         if(pos[u] == 0 && pos[v] == 0) printf("ERROR: %d and %d are not found.\n",u,v);
31         else if(pos[u] == 0||pos[v] == 0) printf("ERROR: %d is not found.\n",pos[u] == 0?u:v);
32         else LCA(1,n,1,u,v);
33     }
34     return 0;
35 }