牛客 打印二叉树边界节点

题目链接：https://www.nowcoder.com/practice/33b88978734c42b68699d0c7cef9b598?tpId=101&tqId=33230&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

　　比较繁琐的 Coding 题，考验基本功。

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] << " 
"[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 << "]" << "
";
    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 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;

struct TreeNode {
    int lch = 0, rch = 0;
};

int N, root;
TreeNode tree[maxN];

// PS:严格空间复杂度O(h)实在不符合我的美学
void printOne() {
    QI Q;
    int cnt = 1;
    VI l, mid, r;

    Q.push(root);
    l.PB(root);
    // 利用BFS求左右边界节点，这里也不满足空间复杂度O(h)的要求
    while(!Q.empty()) {
        int rt = Q.front(); Q.pop();
        --cnt;

        if(tree[rt].lch) Q.push(tree[rt].lch);
        if(tree[rt].rch) Q.push(tree[rt].rch);

        if(cnt == 0) {
            r.PB(rt);
            cnt = Q.size();
            if(!Q.empty()) l.PB(Q.front());
        }
    }

    // 这里用栈代替递归调用，实际上做不到严格空间复杂度O(h)，因为递归调用也要耗空间
    stack< PII > sk;
    sk.push(MP(root, 0)); // 第二维是高度
    // 先序遍历取出所有非左右边界的叶子结点
    while(!sk.empty()) {
        int rt = sk.top().ft, h = sk.top().sd; sk.pop();

        if(!tree[rt].rch && !tree[rt].lch && l[h] != rt && r[h] != rt) mid.PB(rt);

        if(tree[rt].rch) sk.push(MP(tree[rt].rch, h + 1));
        if(tree[rt].lch) sk.push(MP(tree[rt].lch, h + 1));
    }

    Rep(i, l.size()) {
        if(i) printf(" ");
        printf("%d", l[i]);
    }
    Rep(i, mid.size()) printf(" %d", mid[i]);
    rFor(i, r.size() - 1, 1) if(r[i] != l[i]) printf(" %d", r[i]);
    printf("
");
}

void printTwoLeft(int rt) {
    bool flag = true; // 标记左边界是否打印完了
    SKI sk;
    sk.push(rt);
    // 先序遍历
    while(!sk.empty()) {
        int p = sk.top(); sk.pop();

        if(flag || !tree[p].rch && !tree[p].lch) printf(" %d", p);

        if(tree[p].rch) sk.push(tree[p].rch);
        if(tree[p].lch) sk.push(tree[p].lch);

        if(!tree[p].rch && !tree[p].lch) flag = false;
    }
}

void printTwoRight(int rt) {
    bool flag = true; // 标记右边界是否打印完了
    SKI sk;
    VI ret;
    sk.push(rt);
    // 反先序遍历(中右左)
    while(!sk.empty()) {
        int p = sk.top(); sk.pop();

        if(flag || !tree[p].rch && !tree[p].lch) ret.PB(p);

        if(tree[p].lch) sk.push(tree[p].lch);
        if(tree[p].rch) sk.push(tree[p].rch);

        if(!tree[p].rch && !tree[p].lch) flag = false;
    }

    rFor(i, ret.size() - 1, 0) printf(" %d", ret[i]);
}

void printTwo() {
    int rt = root;

    // 找到第一个有两个孩子的节点rt，并沿途打印节点
    while(rt && !(tree[rt].rch && tree[rt].lch)) {
        if(rt != root) printf(" ");
        printf("%d", rt);
        if(tree[rt].rch == 0 && tree[rt].lch == 0) break;    // 不存在拥有两个孩子的节点,直接退出 
        if(tree[rt].rch) rt = tree[rt].rch;
        if(tree[rt].lch) rt = tree[rt].lch;
    }
    if(rt && tree[rt].rch && tree[rt].rch) {
        if(rt != root) printf(" ");
        printf("%d", rt);
    }
    else {
        printf("
");
        return;
    }

    printTwoLeft(tree[rt].lch);
    printTwoRight(tree[rt].rch);
    printf("
");
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    scanf("%d%d", &N, &root);
    Rep(i, N) {
        int fa, lch, rch;
        scanf("%d%d%d", &fa, &lch, &rch);

        tree[fa].lch = lch;
        tree[fa].rch = rch;
    }

    printOne();
    printTwo();
    return 0;
}