【紫书】Tree UVA

题意：给你中序后序 求某叶子节点使得从根到该节点权值和最小。若存在多个，输出其权值最小的那个。

题解：先建树，然后暴力dfs/bfs所有路径，取min

技巧：递归传参数，l1,r1,l2,r2, sum,root,

代码：

#define _CRT_SECURE_NO_WARNINGS
#include "stdio.h"
#include<stdio.h>
#include<algorithm>
#include<string>
#include<vector>
#include<list>
#include<set>
#include<iostream>
#include<string.h>
#include<queue>
#include<string>
#include<sstream>
using namespace std;
const int maxn = 1e4+5;
string line; int n;
int a[maxn],in_order[maxn],post_order[maxn];
int lch[maxn], rch[maxn];
int mn,best;
bool read_list(int *a) {
    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) {//将中序l1,r1 将后序l2,r2 建成一棵树，返回树根
    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 += u;
    if (!lch[u] && !rch[u]) if (sum < mn || (sum == mn&&u < best)) {best = u; mn = sum;}
    if (lch[u])dfs(lch[u], sum);
    if (rch[u])dfs(rch[u], sum);
}
int main() {
    while (read_list(in_order)) {
        read_list(post_order);
        build(0, n - 1, 0, n - 1);
        mn = 1e9;
        dfs(post_order[n - 1], 0);
        cout << best << "
";
    }

    //system("pause");
}