LCA问题： Lowest Common Ancestor

题目链接【传送门】
之前写过一次，用的是空间换时间，也就是前序、中序构建二叉树，求解每个结点的从根结点到此结点的路径（逆序）存入map中，对查询的时候，遍历二者的路径找最远的公共元素作为输出，AC是AC了，但是这个代码量……比较累赘

#include
using namespace std;
int M,N,u,v;
vector<int>pre;
vector<int>in;
struct Node{
	int val;
	Node*left=NULL;
	Node*right=NULL;
	Node*parent=NULL;
};
Node*build(int inL,int inR,int preL,int preR){
	if(inL>inR||preL>preR||inL<0||inR>=N||preL<0||preR>=N){
		return NULL;
	}
	Node* root=new Node;
	
	root->val=pre[preL];
	int k;
	for(k=inL;k<=inR;k++){
		if(in[k]==root->val){
			break;
		}
	}
	int num=k-inL;
	root->left=build(inL,k-1,preL+1,preL+num);
	root->right=build(k+1,inR,preL+num+1,preR);
	return root;
}
map<int,vector<int>>path;
map<int,Node*>nodes;
void levelTravel(Node*root){
	queue<Node*>q;
	q.push(root);
	while(!q.empty()){
		Node*top=q.front();
		nodes[top->val]=top;
		q.pop();
		if(top->left!=NULL){
			top->left->parent=top;
			q.push(top->left);
		}
		if(top->right!=NULL){
			top->right->parent=top;
			q.push(top->right);
		}
	}
}
int main(){
	scanf("%d%d",&M,&N);
	pre.resize(N);
	in.resize(N);
	for(int i=0;i<N;i++){
		scanf("%d",&pre[i]);
	}
	in=pre;
	sort(in.begin(),in.end()); 
	Node*root=build(0,N-1,0,N-1);
	levelTravel(root);
	for(int i=0;i<M;i++){
		scanf("%d%d",&u,&v);
		if(find(pre.begin(),pre.end(),u)==pre.end()){
			if(find(pre.begin(),pre.end(),v)==pre.end()){
				printf("ERROR: %d and %d are not found.\n",u,v);
			}else{
				printf("ERROR: %d is not found.\n",u);
			}
		}else{
			if(find(pre.begin(),pre.end(),v)==pre.end()){
				printf("ERROR: %d is not found.\n",v);
			}else{
				//寻找两个结点的最近的父辈
				int p;
				if(path.find(u)==path.end()){
					p=u;
					while(nodes[p]->parent!=NULL){
						path[u].push_back(p);
						p=nodes[p]->parent->val;
					}
					reverse(path[u].begin(),path[u].end());
				}
				if(path.find(v)==path.end()){
					p=v;
					while(nodes[p]->parent!=NULL){
						path[v].push_back(p);
						p=nodes[p]->parent->val;
					}
					reverse(path[v].begin(),path[v].end());
				}
				int a=root->val;
				for(int i=0;i<min(path[u].size(),path[v].size());i++){
					if(path[u][i]==path[v][i]){
						a=path[u][i];
					}else{
						break;
					}
				}
				if(a==u){
					printf("%d is an ancestor of %d.\n",a,v);
				}else if(a==v){
					printf("%d is an ancestor of %d.\n",a,u);
				}else{
					printf("LCA of %d and %d is %d.\n",u,v,a);
				}
			}
		}
	}
	return 0;
}

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今天重写的时候发现LCA可以不用这么复杂，于是参考【LCA in Binary Trees柳神】的思路用递归的方式重构了一次……结果喜提超时了一个测试点24/30：

#include
using namespace std;
int M,N,u,v;
vector<int>pre,in;
map<int,int>inpos;
map<int,bool>mp;
void LCA(int preL,int preR,int inL,int inR){
	int rootPos=preL;
	if(u<pre[rootPos]&&v>pre[rootPos]||u>pre[rootPos]&&v<pre[rootPos]){
		printf("LCA of %d and %d is %d.\n",u,v,pre[rootPos]);
	}else if(u<pre[rootPos]&&u<pre[rootPos]){
		//左子树 
		int k=inpos[pre[rootPos]];
		int num=k-inL;
		LCA(preL+1,preL+num,inL,k-1);
	}else if(u>pre[rootPos]&&u>pre[rootPos]){
		//右子树
		int k=inpos[pre[rootPos]];
		int num=k-inL;
		LCA(preL+1+num,preR,k+1,inR); 
	}else if(u==pre[rootPos]){
		printf("%d is an ancestor of %d.\n",u,v);
	}else if(v==pre[rootPos]){
		printf("%d is an ancestor of %d.\n",v,u);
	}
}
int main(){
	scanf("%d%d",&M,&N);
	pre.resize(N);
	for(int i=0;i<N;i++){
		scanf("%d",&pre[i]);
		mp[pre[i]]=true;
	}
	in=pre;
	sort(in.begin(),in.end());
	for(int i=0;i<N;i++){
		inpos[in[i]]=i;
	}
	while(M--){
		scanf("%d%d",&u,&v);
		if(mp.find(u)==mp.end()){
			if(mp.find(v)==mp.end()){
				printf("ERROR: %d and %d are not found.\n",u,v);
			}else{
				printf("ERROR: %d is not found.\n",u);
			}
		}else{
			if(mp.find(v)==mp.end()){
				printf("ERROR: %d is not found.\n",v);
			}else{
				//找到最近公共结点
				LCA(0,N-1,0,N-1); 
			}
		}
	}
	
	return 0;
}
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看了柳神的此题题解，发现……还是自己对BST树了解有待提高。思路是首先用map存储所有key,对输入先判断是不是树中的结点；接着遍历pre数组，由二叉排序树的性质可知，当顺序遍历pre数组时，第一个出现pre[index]<=u&&pre[index]>=v||pre[index]>=u&&pre[index]<=v即为答案结点，按照要求输出即可

AC代码如下（代码量和效率的双赢了属实，但是确实对性质理解不到位很难想到这个解法）：

#include
using namespace std;
int N,M,a,b;
vector<int>pre;
map<int,bool>mp;
int main(){
	scanf("%d%d",&M,&N);
	pre.resize(N);
	for(int i=0;i<N;i++){
		scanf("%d",&pre[i]);
		mp[pre[i]]=true;
	}
	while(M--){
		scanf("%d%d",&a,&b);
		if(mp.find(a)==mp.end()){
			if(mp.find(b)==mp.end()){
				printf("ERROR: %d and %d are not found.\n",a,b);
			}else{
				printf("ERROR: %d is not found.\n",a);	
			}
		}else{
			if(mp.find(b)==mp.end()){
				printf("ERROR: %d is not found.\n",b);
			}else{
				int ans;
				for(int i=0;i<N;i++){
					if(pre[i]<a&&pre[i]>b||pre[i]>a&&pre[i]<b||pre[i]==a||pre[i]==b){
						ans=pre[i];
						break;
					}
				}
				if(ans!=a){
					if(ans!=b){
						printf("LCA of %d and %d is %d.\n",a,b,ans);
					}else{
						printf("%d is an ancestor of %d.\n",ans,a);
					}
				}else{
					printf("%d is an ancestor of %d.\n",ans,b);
				}
			}
		}
	}
	return 0;
} 
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原题如下：
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
1
2
3
4
5
6
7
8

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.
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