UVA-548 Tree

2016年05月04日 |

acm

| acm

题目：

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

代码：

 // oj.cpp : 定义控制台应用程序的入口点。
//
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <numeric>

using namespace std;

int I[10010];
int P[10010];

int ans;
int minp;

void solve(int beg, int end, int left, int path)
{
    if (beg == end)
        return;
    if (end - beg <= 1)
    {
        if (I[beg] + path < minp)
        {
            minp = I[beg] + path;
            ans = I[beg];
        }
        return;
    }
    int i = find(I + beg, I + end, P[left]) - I;
    solve(i + 1, end, left - 1, path + P[left]);
    solve(beg, i, left - end + i, path + P[left]);
}

int main()
{
    ios::sync_with_stdio(false);
    string L, M;
    while (getline(cin, L))
    {
        getline(cin, M);
        minp = numeric_limits<int>::max();
        istringstream stL(L), stM(M);
        int N = 0;
        while (stL >> I[N])
            stM >> P[N++];
        solve(0, N, N - 1, 0);
        cout << ans << endl;
    }
    return 0;
}

解析&吐槽：

通过中序和后序遍历重建二叉树。需要注意的是题目要求输出“最短的一条路径” 的叶子节点的值。也就是说需要把分支节点的值也考虑在内。