Question：
You are to determine the value of the leaf node in a given binary tree that is the terminal node of a
path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values
of nodes along that path.
Input
The input file will contain a description of the binary tree given as the inorder and postorder traversal
sequences of that tree. Your program will read two line (until end of file) from the input file. The first
line will contain the sequence of values associated with an inorder traversal of the tree and the second
line will contain the sequence of values associated with a postorder traversal of the tree. All values
will be different, greater than zero and less than 10000. You may assume that no binary tree will have
more than 10000 nodes or less than 1 node.
Output
For each tree description you should output the value of the leaf node of a path of least value. In the
case of multiple paths of least value you should pick the one with the least value on the terminal node.
Sample Input
3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255
Sample Output
1
3
255
題目大意：給你一棵二叉樹的中序遍歷和後序遍歷，找到一個葉子節點使這條路權值和最小，如果有多解，則輸出節點最小的葉節點

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <sstream>  //為讀取輸入
using namespace std;
const int maxn=1e4+5;
const int INF=0x3f3f3f3f
 
;
int l[maxn],r[maxn],a[maxn],b[maxn];
int n,best_sum,best;
int build(int L1,int R1,int L2,int R2)
{
    if(L1>R1)
        return 0;
    int root=b[R2];
    int p=L1;
    while(a[p]!=root)
        p++;
    int cnt=p-L1;
    l[root]=build(L1,p-1,L2,L2+cnt-1);
    r[root]=build(p+1,R1,L2+cnt,R2-1);
    return
 
 root;
}  //運用遞迴建樹
void dfs(int u,int sum)
{
    sum+=u;
    if(!l[u]&&!r[u])
    {
        if(sum<best_sum||(sum==best_sum&&u<best))
        {
            best=u;
            best_sum=sum;
        }
    }
        if(l[u])   dfs(l[u],sum);   
        if(r[u])   dfs(r[u],sum);
}
bool read(int *w)  //輸入讀取（這十分重要）
{
    string line;
    if(!getline(cin,line))
        return false;
    stringstream ss(line);
    n=0;
    int x;
    while(ss>>x)
        w[n++]=x;  
    return n>0;
}
int main()
{
    while (read(a))  //讀取後分別儲存在陣列a和陣列b中
    {
        read(b);
        best_sum=INF;
        build(0,n-1,0,n-1);
        dfs(b[n-1],0);
        printf("%d\n",best);
    }
    return 0;
}

體會：基本建樹一定要掌握，這是一個二叉樹建樹的基本題型