数据结构 有序树转二叉树 (树的遍历)

Description

计算输入有序树的深度和有序树转化为二叉树之后树的深度。

Input

输入包含多组数据。每组数据第一行为一个整数n（2<=n<=30000）代表节点的数量，接下来n-1行，两个整数a、b代表a是b的父亲结点。

Output

输出当前树的深度和转化成二叉树之后的深度。

Sample Input

5

1 5

1 3

5 2

1 4

Sample Output

3 4

HINT

Append Code

析：这个题很好说，只要遍历一次就能得到答案，由于要先有序树转成二叉树，我也没听过，我百度了一下怎么转，看到百度百科中有，我总结一下就是，左儿子，

右兄弟，和那个字典树有的一拼，也就是左结点是儿子结点，而右结点是兄弟结点，那么我们可以用两个参数来记录深度，二叉树既然是右兄弟，那么同一深度的就会多一层，

加1，这样最大的就是答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e4 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
bool in[maxn];
int ans1, ans2;
 
void dfs(int u, int cnt1, int cnt2){
    ans1 = Max(ans1, cnt1);
    ans2 = Max(ans2, cnt2);
    for(int i = 0; i < G[u].size(); ++i)
        dfs(G[u][i], cnt1+1, cnt2+i+1);
}
 
int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 0; i <= n; ++i)  G[i].clear(), in[i] = 0;
        int u, v;
        for(int i = 1; i < n; ++i){
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            in[v] = true;
        }
        ans1 = ans2 = 0;
        for(int i = 1; i <= n; ++i) if(!in[i]){
            dfs(i, 1, 1);  break;
        }
        printf("%d %d\n", ans1, ans2);
    }
    return 0;
}