给出二叉树的中后序遍历 建立二叉树 并找出权和最小值

Dfs类型题

收获：明确二叉树的三种优先遍历，以及用c++输入流输出流的使用

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<sstream>
#include<iostream>

using namespace std;
const int MaxN = 1e4;
int in_order[MaxN + 5], post_order[MaxN + 5];
int lch[MaxN+ 5], rch[MaxN + 5];
int n;
int best, best_sum = 1<<30;

bool read_list(int *a)
{
    string line;
    if(!getline(cin, line)) return false;
    stringstream ss(line);
    n = 0;
    int x;
    while(ss >> x) a[n++] = x;
    return n > 0;
}

int Build(int L1, int R1, int L2, int R2)
{
    if(L1 > R1)return 0;
    int root = post_order[R2];
    int p = L1;
    while(in_order[p] != root) p++;
    int cnt = p - L1;
    lch[root] = Build(L1, p - 1, L2, L2 + cnt - 1);
    rch[root] = Build(p + 1, R1, L2 + cnt, R2 - 1);
    return root;
}
void Dfs(int u, int sum)
{
    sum = sum + u;
    if(!lch[u] && !rch[u])
    {
        if(sum < best_sum || (sum == best_sum && u < best))
        {
            best = u;
            best_sum = sum;
        }
    }
    if(lch[u]) Dfs(lch[u], sum);
    if(rch[u]) Dfs(rch[u], sum);
}

int main()
{
    while(read_list(in_order))
    {
        best_sum = 1<<30;
        read_list(post_order);
        Build(0, n - 1, 0, n - 1);
        Dfs(post_order[n - 1], 0);
        cout << best << "
";
    }
    return 0;
}