题目大意
略。
分析
比较繁琐的 Coding 题，考验基本功。
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] << " \n"[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 << "]" << "\n";
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 // PS:严格空间复杂度O(h)实在不符合我的美学
147 void printOne() {
148 QI Q;
149 int cnt = 1;
150 VI l, mid, r;
151
152 Q.push(root);
153 l.PB(root);
154 // 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
155 while(!Q.empty()) {
156 int rt = Q.front(); Q.pop();
157 --cnt;
158
159 if(tree[rt].lch) Q.push(tree[rt].lch);
160 if(tree[rt].rch) Q.push(tree[rt].rch);
161
162 if(cnt == 0) {
163 r.PB(rt);
164 cnt = Q.size();
165 if(!Q.empty()) l.PB(Q.front());
166 }
167 }
168
169 // 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
170 stack< PII > sk;
171 sk.push(MP(root, 0)); // 第二维是高度
172 // 先序遍历取出所有非左右边界的叶子结点
173 while(!sk.empty()) {
174 int rt = sk.top().ft, h = sk.top().sd; sk.pop();
175
176 if(!tree[rt].rch && !tree[rt].lch && l[h] != rt && r[h] != rt) mid.PB(rt);
177
178 if(tree[rt].rch) sk.push(MP(tree[rt].rch, h + 1));
179 if(tree[rt].lch) sk.push(MP(tree[rt].lch, h + 1));
180 }
181
182 Rep(i, l.size()) {
183 if(i) printf(" ");
184 printf("%d", l[i]);
185 }
186 Rep(i, mid.size()) printf(" %d", mid[i]);
187 rFor(i, r.size() - 1, 1) if(r[i] != l[i]) printf(" %d", r[i]);
188 printf("\n");
189 }
190
191 void printTwoLeft(int rt) {
192 bool flag = true; // 标记左边界是否打印完了
193 SKI sk;
194 sk.push(rt);
195 // 先序遍历
196 while(!sk.empty()) {
197 int p = sk.top(); sk.pop();
198
199 if(flag || !tree[p].rch && !tree[p].lch) printf(" %d", p);
200
201 if(tree[p].rch) sk.push(tree[p].rch);
202 if(tree[p].lch) sk.push(tree[p].lch);
203
204 if(!tree[p].rch && !tree[p].lch) flag = false;
205 }
206 }
207
208 void printTwoRight(int rt) {
209 bool flag = true; // 标记右边界是否打印完了
210 SKI sk;
211 VI ret;
212 sk.push(rt);
213 // 反先序遍历(中右左)
214 while(!sk.empty()) {
215 int p = sk.top(); sk.pop();
216
217 if(flag || !tree[p].rch && !tree[p].lch) ret.PB(p);
218
219 if(tree[p].lch) sk.push(tree[p].lch);
220 if(tree[p].rch) sk.push(tree[p].rch);
221
222 if(!tree[p].rch && !tree[p].lch) flag = false;
223 }
224
225 rFor(i, ret.size() - 1, 0) printf(" %d", ret[i]);
226 }
227
228 void printTwo() {
229 int rt = root;
230
231 // 找到第一个有两个孩子的节点rt，并沿途打印节点
232 while(rt && !(tree[rt].rch && tree[rt].lch)) {
233 if(rt != root) printf(" ");
234 printf("%d", rt);
235 if(tree[rt].rch) rt = tree[rt].rch;
236 if(tree[rt].lch) rt = tree[rt].lch;
237 }
238 if(rt && tree[rt].rch && tree[rt].rch) {
239 if(rt != root) printf(" ");
240 printf("%d", rt);
241 }
242 else {
243 printf("\n");
244 return;
245 }
246
247 printTwoLeft(tree[rt].lch);
248 printTwoRight(tree[rt].rch);
249 printf("\n");
250 }
251
252 int main(){
253 //freopen("MyOutput.txt","w",stdout);
254 //freopen("input.txt","r",stdin);
255 //INIT();
256 scanf("%d%d", &N, &root);
257 Rep(i, N) {
258 int fa, lch, rch;
259 scanf("%d%d%d", &fa, &lch, &rch);
260
261 tree[fa].lch = lch;
262 tree[fa].rch = rch;
263 }
264
265 printOne();
266 printTwo();
267 return 0;
268 }
来源：oschina
链接：https://my.oschina.net/u/4264305/blog/4289354