牛客 统计和生成所有不同的二叉树

题目链接：https://www.nowcoder.com/practice/4869f80e29e94d49a20f8d54d3bf2a65?tpId=101&tqId=33250&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking

题目大意

　　略。

分析

　　关于为什么是卡特兰数，可以这么想，对于经典问题“N个数入栈，有多少种出栈顺序？”，这个问题的答案是卡特兰数的第 N 项，那么本题就可以这么想，“N个数出栈，有多少种入栈顺序？”，这是因为中序遍历就是递归，就是要入栈出栈的。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
#define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x3f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
    Rep(i, v.size()) out << v[i] << " \n"[i == v.size() - 1];
    return out;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "\n";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}

typedef unsigned int uI;
typedef long long LL;
typedef unsigned long long uLL;
typedef vector< int > VI;
typedef vector< bool > VB;
typedef vector< char > VC;
typedef vector< double > VD;
typedef vector< string > VS;
typedef vector< LL > VL;
typedef vector< VI > VVI;
typedef vector< VB > VVB;
typedef vector< VS > VVS;
typedef vector< VL > VVL;
typedef vector< VVI > VVVI;
typedef vector< VVL > VVVL;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef pair< int, string > PIS;
typedef pair< string, int > PSI;
typedef pair< string, string > PSS;
typedef pair< double, double > PDD;
typedef vector< PII > VPII;
typedef vector< PLL > VPLL;
typedef vector< VPII > VVPII;
typedef vector< VPLL > VVPLL;
typedef vector< VS > VVS;
typedef map< int, int > MII;
typedef unordered_map< int, int > uMII;
typedef map< LL, LL > MLL;
typedef map< string, int > MSI;
typedef map< int, string > MIS;
typedef set< int > SI;
typedef stack< int > SKI;
typedef deque< int > DQI;
typedef queue< int > QI;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e6 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

LL N; 

LL fac[maxN << 1];
void init_fact() {
    fac[0] = 1;
    For(i, 1, (N << 1)) {
        fac[i] = (i * fac[i - 1]) % mod;
    }
}

//ax + by = gcd(a, b) = d
// 扩展欧几里德算法
/**
 *    a*x + b*y = 1
 *    如果ab互质，有解
 *    x就是a关于b的逆元
 *    y就是b关于a的逆元
 *     
 *    证明： 
 *        a*x % b + b*y % b = 1 % b
 *        a*x % b = 1 % b
 *        a*x = 1 (mod b)
 */
inline void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
    if (!b) {d = a, x = 1, y = 0;}
    else{
        ex_gcd(b, a % b, y, x, d);
        y -= x * (a / b);
    }
}

// 求a关于p的逆元，如果不存在，返回-1 
// a与p互质，逆元才存在 
inline LL inv_mod(LL a, LL p = mod){
    LL d, x, y;
    ex_gcd(a, p, x, y, d);
    return d == 1 ? (x % p + p) % p : -1;
}

inline LL comb_mod(LL m, LL n) {
    LL ret;

    if(m > n) swap(m, n);
    
    ret = (fac[n] * inv_mod(fac[m], mod)) % mod;
    ret = (ret * inv_mod(fac[n - m], mod)) % mod;
    
    return ret;
}

// 求卡特兰数 
inline LL Catalan(LL x) {
    if(x == 0) return 1;
    LL ret = inv_mod(x + 1, mod) * comb_mod(x, 2 * x);
    return ret % mod;
} 

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    scanf("%lld", &N);
    init_fact();
    if(N < 1) N = 0;
    printf("%lld\n", Catalan(N));
    return 0;
}