二叉搜尋樹的建樹和尋找最近公共祖先（題目給出了BST的前序遍歷，而前序遍歷升序排列就是BST的中序遍歷了）

#include <bits/stdc++.h>
using namespace std;

typedef struct node *Node;
typedef struct node{
    int val;
    Node l,r;
}node;

Node T;
int M,N;
int pre[10001];

Node buildTree(int s,int e)
{
    if(e < s) return NULL;
    if(s == e){
        Node tnode = (Node)malloc(sizeof(node));
        tnode->val = pre[s];
        tnode->l = NULL;
        tnode->r = NULL;
        return tnode;
    }
    int i = s+1;
    while(i<=e && pre[i]<pre[s])
        i++;
    Node tnode = (Node)malloc(sizeof(node));
    tnode->val = pre[s];
    tnode->l = buildTree(s+1,i-1);
    tnode->r = buildTree(i,e);
    return tnode;
}

bool findd(Node tree, int a, int b,int f)
{
    if(f == 0){
        if(tree == NULL){
            printf("ERROR: %d and %d are not found.\n", a, b);
			return true;
        }
        if(a<tree->val && b<tree->val)return findd(tree->l,a,b,0);
        else if(a>tree->val && b>tree->val) return findd(tree->r,a,b,0);
        else{
            bool fda = tree->val == a || findd(tree->l,a,0,1) ||  findd(tree->r,a,0,1);
            bool fdb = tree->val == b || findd(tree->l,b,0,1) ||  findd(tree->r,b,0,1);
            if(fda&&fdb){
                if(tree->val == a||tree->val == b)
                    printf("%d is an ancestor of %d.\n", tree->val == a ? a : b, tree->val == b ? a : b);
                else
                    printf("LCA of %d and %d is %d.\n", a, b, tree->val);
            }
            else if (fda == false && fdb == false) {
				printf("ERROR: %d and %d are not found.\n", a, b);
			}
			else {
				printf("ERROR: %d is not found.\n", fda ? b : a);
			}
			return true;
        }
    }
    else{
        if(tree == NULL) return false;
        if(tree->val == a) return true;
        if(tree->val > a)return findd(tree->l,a,0,1);
        return findd(tree->r,a,0,1);
    }

}

int main()
{
    cin>>M>>N;
    int a,b;
    for(int i=0;i<N;i++){
        cin>>pre[i];
    }
    T = buildTree(0,N-1);
    for(int i=0;i<M;i++){
        cin>>a>>b;
        findd(T, a, b, 0);
    }

    return 0;
}
 

看到有一種非常巧妙的方法，都不需要建樹：

map<int, bool> mp用來標記樹中所有出現過的結點，遍歷一遍pre陣列，將當前結點標記為a，如果u和v分別在a的左、右，或者u、v其中一個就是當前a，即(a >= u && a <= v) || (a >= v && a <= u)，說明找到了這個共同最低祖先a，退出當前迴圈，最後根據要求輸出結果即可

#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;
}