RT，方法是找lca再用dfs序分别判断两个lca是否出现在另外一条路径上

测试结果

代码：

#include<bits/stdc++.h>
#define N 100005
using namespace std;
inline int read()
{
	int x=0; char ch=getchar();
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return x;
}

int n;
int fa[N][22],dep[N],lg[N],size[N],dfn[N],tim;
vector<int>e[N];

inline void dfs(int x,int f)
{
	dfn[x]=++tim;
	fa[x][0]=f,dep[x]=dep[f]+1,size[x]=1;
	for(int i=1;i<lg[dep[x]];++i) fa[x][i]=fa[fa[x][i-1]][i-1];
	for(int i=0;i<e[x].size();++i){
		int t=e[x][i];
		if(t==f) continue;
		dfs(t,x);
		size[x]+=size[t];
	}
}

inline int lca(int x,int y)
{
	if(dep[x]<dep[y]) swap(x,y);
	while(dep[x]>dep[y])
		x=fa[x][lg[dep[x]-dep[y]]-1];
	if(x==y) return x;
	for(int i=lg[dep[x]]-1;i>=0;--i)
		if(fa[x][i]!=fa[y][i]) 
			x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}

inline bool solve(int x1,int y1,int x2,int y2)
{
	int l1=lca(x1,y1),l2=lca(x2,y2);
	if(dep[x1]>=dep[l2]&&dfn[x1]<=dfn[l2]+size[l2]-1&&dep[l2]>=dep[l1]&&dfn[l2]<=dfn[l1]+size[l1]-1) return 1;
	if(dep[y1]>=dep[l2]&&dfn[y1]<=dfn[l2]+size[l2]-1&&dep[l2]>=dep[l1]&&dfn[l2]<=dfn[l1]+size[l1]-1) return 1;
	if(dep[x2]>=dep[l1]&&dfn[x2]<=dfn[l1]+size[l1]-1&&dep[l1]>=dep[l2]&&dfn[l1]<=dfn[l2]+size[l2]-1) return 1;
	if(dep[y2]>=dep[l1]&&dfn[y2]<=dfn[l1]+size[l1]-1&&dep[l1]>=dep[l2]&&dfn[l1]<=dfn[l2]+size[l2]-1) return 1;
	return 0;
}

int main()
{
	n=read(); int q=read();
	for(int i=1;i<n;++i){
		int x=read(),y=read();
		e[x].push_back(y),e[y].push_back(x);
	}
	for(int i=1;i<=n;++i) lg[i]=lg[i-1]+((1<<lg[i-1])==i);
	dfs(1,1);
	for(int i=1;i<=q;++i){
		int x1=read(),y1=read(),x2=read(),y2=read();
		if(solve(x1,y1,x2,y2)) putchar('Y'),putchar('\n');
		else putchar('N'),putchar('\n');
	}	
	return 0;
}