牛客打印二叉树边界节点

灰太狼 2023-06-04 14:53 30阅读 0赞

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][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 ###

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

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 }

转载于:https://www.cnblogs.com/zaq19970105/p/11355930.html

[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]: 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
[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
[ExpandedBlockStart.gif]: /images/20230601/acf140d61b4440c4afd7994c2418fb88.png