题目大意
略。
分析
线索二叉树 Morris 遍历。
获取反转的右边界的时候我还是用了栈，比较方便，实际上应该反转两次链表，这样才能做到空间复杂度 O(1)。
代码如下
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i) 6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i) 7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 24 25 #define ms0(a) memset(a,0,sizeof(a)) 26 #define msI(a) memset(a,0x3f,sizeof(a)) 27 #define msM(a) memset(a,-1,sizeof(a)) 28 29 #define MP make_pair 30 #define PB push_back 31 #define ft first 32 #define sd second 33 34 template<typename T1, typename T2> 35 istream &operator>>(istream &in, pair<T1, T2> &p) { 36 in >> p.first >> p.second; 37 return in; 38 } 39 40 template<typename T> 41 istream &operator>>(istream &in, vector<T> &v) { 42 for (auto &x: v) 43 in >> x; 44 return in; 45 } 46 47 template<typename T> 48 ostream &operator<<(ostream &out, vector<T> &v) { 49 Rep(i, v.size()) out << v[i] << " "[i == v.size() - 1]; 50 return out; 51 } 52 53 template<typename T1, typename T2> 54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 55 out << "[" << p.first << ", " << p.second << "]" << " "; 56 return out; 57 } 58 59 inline int gc(){ 60 static const int BUF = 1e7; 61 static char buf[BUF], *bg = buf + BUF, *ed = bg; 62 63 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 64 return *bg++; 65 } 66 67 inline int ri(){ 68 int x = 0, f = 1, c = gc(); 69 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 70 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 71 return x*f; 72 } 73 74 template<class T> 75 inline string toString(T x) { 76 ostringstream sout; 77 sout << x; 78 return sout.str(); 79 } 80 81 inline int toInt(string s) { 82 int v; 83 istringstream sin(s); 84 sin >> v; 85 return v; 86 } 87 88 //min <= aim <= max 89 template<typename T> 90 inline bool BETWEEN(const T aim, const T min, const T max) { 91 return min <= aim && aim <= max; 92 } 93 94 typedef long long LL; 95 typedef unsigned long long uLL; 96 typedef vector< int > VI; 97 typedef vector< bool > VB; 98 typedef vector< char > VC; 99 typedef vector< double > VD; 100 typedef vector< string > VS; 101 typedef vector< LL > VL; 102 typedef vector< VI > VVI; 103 typedef vector< VB > VVB; 104 typedef vector< VS > VVS; 105 typedef vector< VL > VVL; 106 typedef vector< VVI > VVVI; 107 typedef vector< VVL > VVVL; 108 typedef pair< int, int > PII; 109 typedef pair< LL, LL > PLL; 110 typedef pair< int, string > PIS; 111 typedef pair< string, int > PSI; 112 typedef pair< string, string > PSS; 113 typedef pair< double, double > PDD; 114 typedef vector< PII > VPII; 115 typedef vector< PLL > VPLL; 116 typedef vector< VPII > VVPII; 117 typedef vector< VPLL > VVPLL; 118 typedef vector< VS > VVS; 119 typedef map< int, int > MII; 120 typedef unordered_map< int, int > uMII; 121 typedef map< LL, LL > MLL; 122 typedef map< string, int > MSI; 123 typedef map< int, string > MIS; 124 typedef set< int > SI; 125 typedef stack< int > SKI; 126 typedef deque< int > DQI; 127 typedef queue< int > QI; 128 typedef priority_queue< int > PQIMax; 129 typedef priority_queue< int, VI, greater< int > > PQIMin; 130 const double EPS = 1e-8; 131 const LL inf = 0x7fffffff; 132 const LL infLL = 0x7fffffffffffffffLL; 133 const LL mod = 1e9 + 7; 134 const int maxN = 1e6 + 7; 135 const LL ONE = 1; 136 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 137 const LL oddBits = 0x5555555555555555; 138 139 struct TreeNode { 140 int lch = 0, rch = 0; 141 }; 142 143 int N, root; 144 TreeNode tree[maxN]; 145 146 // 返回搜索二叉树最大值节点的序号 147 int getMax(int rt, int cur) { 148 while(tree[rt].rch && tree[rt].rch != cur) rt = tree[rt].rch; 149 return rt; 150 } 151 152 void getReverseRightEdge(int rt,VI &ret) { 153 SKI sk; 154 while(rt) { 155 sk.push(rt); 156 rt = tree[rt].rch; 157 } 158 while(!sk.empty()) { 159 ret.PB(sk.top()); 160 sk.pop(); 161 } 162 } 163 164 // mode = 1:先序 165 // mode = 2:中序 166 // mode = 3:后序 167 void Morris(int rt, int mode) { 168 if(rt == 0) return; 169 int cur = rt; 170 int mostRight = -1; 171 VI ret; 172 173 while(cur) { 174 if(tree[cur].lch) { 175 mostRight = getMax(tree[cur].lch, cur); 176 177 if(tree[mostRight].rch == cur) { // 第二次来到cur 178 tree[mostRight].rch = 0; 179 if(mode == 2) ret.PB(cur); 180 if(mode == 3) getReverseRightEdge(tree[cur].lch, ret); 181 } 182 else { // 第一次来到cur 183 if(mode == 1) ret.PB(cur); 184 tree[mostRight].rch = cur; 185 cur = tree[cur].lch; 186 continue; 187 } 188 } 189 else if(mode != 3) ret.PB(cur); 190 cur = tree[cur].rch; 191 } 192 if(mode == 3) getReverseRightEdge(rt, ret); 193 194 Rep(i, ret.size()) printf("%d%c", ret[i], " "[i == ret.size() - 1]); 195 } 196 197 int main(){ 198 //freopen("MyOutput.txt","w",stdout); 199 //freopen("input.txt","r",stdin); 200 //INIT(); 201 scanf("%d%d", &N, &root); 202 Rep(i, N) { 203 int fa, lch, rch; 204 scanf("%d%d%d", &fa, &lch, &rch); 205 206 tree[fa].lch = lch; 207 tree[fa].rch = rch; 208 } 209 210 Morris(root, 1); 211 Morris(root, 2); 212 Morris(root, 3); 213 return 0; 214 }