1143 Lowest Common Ancestor (30 分)

题目大意：给一个前序遍历的二叉搜索树，求两点的最近公共祖先。

二叉搜索树 左<根，根<=右，所以中序遍历一定是升序的。

1.老套路，先建树，然后再找最近公共祖先

2.不用建树，根据性质直接遍历一遍前序遍历，因为中序遍历是有序。(左根右）

建树的代码

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
// 前序二叉搜索树，中序遍历就是从小到大排序后的结果
int n,m;
unordered_map<int,int>mp;
struct node
{
	int v;
	node *left,*right;
	node(int x) : v(x),left(NULL),right(NULL){}
}*tr;
node *build(int preL,int preR,vector<int> &v) //二叉搜索树建树
{
	if(preL > preR) return NULL;
	node *t = new node(v[preL]);
	int k = preL;
	while(v[preL] >= v[k]) k ++; //找到第一个右
	t -> left = build(preL + 1, k - 1,v);
	t -> right = build(k,preR,v);
	return t;
}
int lca(node *t,int a,int b)
{   
	int val = t -> v;
	if(val > a && val > b) return lca(t ->left,a,b);
	else if (val < a && val < b) return lca(t-> right,a,b);
	else return val;
}
int main()
{
	cin >> n >> m;
	vector<int>v(m);
	for(int i = 0; i < m; i ++)
	{
		cin >> v[i];
		mp[v[i]] = 1;
	}
	tr = build(0,m - 1,v);
	while( n --)
	{
		int a,b;
		cin >> a >> b;
		if(!mp.count(a) && mp.count(b)) printf("ERROR: %d is not found.\n",a);
		else if(mp.count(a) && !mp.count(b)) printf("ERROR: %d is not found.\n",b);
		else if (!mp.count(a) && !mp.count(b)) printf("ERROR: %d and %d are not found.\n",a,b);
		else 
		{
			int t = lca(tr,a,b);
			if(t == a) printf("%d is an ancestor of %d.\n",a,b);
			else if(t == b) printf("%d is an ancestor of %d.\n",b,a);
			else printf("LCA of %d and %d is %d.\n",a,b,t);
		}
	}
	return 0;
}

不用建树直接遍历

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;
// 前序二叉搜索树，中序遍历就是从小到大排序后的结果
int n,m,a,b;
unordered_map<int,int>mp;
int main()
{
	cin >> n >> m;
	vector<int>v(m);
	for(int i = 0; i < m; i ++)
	{
		cin >> v[i];mp[v[i]] = 1;
	}
	while( n --)
	{
		cin >> a >> b;
		if(!mp.count(a) && mp.count(b)) printf("ERROR: %d is not found.\n",a);
		else if(mp.count(a) && !mp.count(b)) printf("ERROR: %d is not found.\n",b);
		else if (!mp.count(a) && !mp.count(b)) printf("ERROR: %d and %d are not found.\n",a,b);
		else 
		{
			for(int i = 0; i < m; i ++)
			{   
				if(a < v[i] && v[i] < b || a > v[i] && v[i] > b)
				{   
					printf("LCA of %d and %d is %d.\n",a,b,v[i]);
					break;
				}
				else if(a == v[i]) 
				{
					printf("%d is an ancestor of %d.\n",a,b);break;
				}
				else if(b == v[i]){
					printf("%d is an ancestor of %d.\n",b,a);break;
				}
			}
		}
	}
	return 0;
}