uva 548 中序后序遍历构造树 数组二叉树 stringstream格式转换

Tree
Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

Description
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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

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>

using namespace std;

int inOrder[10005],postOrder[10005],lch[10005],rch[10005];
int n;

int getList(int *a)
{
    string line;
    if(!getline(cin,line))//c++用getline而不用gets（）
        return 0;
    stringstream ss(line);//stringstream常用来进行格式之间的转换，或者用于一行一行的读取。 还可以ss<<line；如果多次使用需要clear()
    n = 0;
    int x;
    while(ss>>x)
        a[n++] = x;// a[n++]这种表示方法给数组赋值，易读性高

    return n > 0;//当字符串为空时返回false，中断读入
}

int build(int L1,int R1,int L2,int R2)
{
    if(L1 > R1)
        return 0;
    int root = postOrder[R2];//找到需求值
    int p = L1;//建立指引
    while(inOrder[p] != root)
        p++;
    int cnt = p-L1;
    lch[root] = build(L1,p-1,L2,L2+cnt-1);//根节点在中间
    rch[root] = build(p+1,R2,L2+cnt,R2-1);//根节点在末尾

    return root;//先遍历，后赋值
}

int best,best_sum;

void dfs(int u,int sum)
{
    sum += u;
    if(!lch[u] && !rch[u])//题目要求
        if(sum < best_sum || (sum == best_sum && u < best))
        {
            best = u;
            best_sum = sum;
        }
    if(lch[u])
        dfs(lch[u],sum);//dfs的根基，若存在，则遍历，二叉树的数组表示。键和值形成节点后下一个节点的对应。
    if(rch[u])
        dfs(rch[u],sum);
}

int main()
{
    //freopen("F:\\data.txt","r",stdin);
    ios::sync_with_stdio(false);

    while(getList(inOrder))
    {
        getList(postOrder);
        build(0,n-1,0,n-1);
        best_sum = 1e+9;
        dfs(postOrder[n-1],0);
        cout<<best<<"\n";
    }

    return 0;
}