548 - Tree

You are todetermine the value of the leaf node in a given binary tree that is theterminal node of a path of least value from the root of the binary tree to anyleaf. The value of a path is the sum of values of nodes along that path.

Input 

The input filewill contain a description of the binary tree given as the inorder andpostorder traversal sequences of that tree. Your program will read two line(until end of file) from the input file. The first line will contain thesequence of values associated with an inorder traversal of the tree and thesecond line will contain the sequence of values associated with a postordertraversal of the tree. All values will be different, greater than zero and lessthan 10000. You may assume that no binary tree will have more than 10000 nodesor less than 1 node.

Output 

For each treedescription you should output the value of the leaf node of a path of leastvalue. In the case of multiple paths of least value you should pick the onewith 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 1218

8 3 11 7 16 18 125

255

255

Sample Output 

1

3

255

代码:

/*

题意：给一棵点带权（权各不相同，都是正整数）二叉树的中序和后序遍历，

找一个叶子使得它到根的路径上的权和最小。如果有多解，该叶子本身的权应尽量小

算法：递归建树，然后DFS。

注意，直接递归求结果也可以，但是先建树的方法不仅直观，而且更好调试

*/

#include<iostream>

#include<string>

#include<sstream>

#include<algorithm>

using namespacestd;

 

//因为各个结点的权值各不相同且都是正整数，直接用权值作为结点编号

const int maxv =10000 + 10;

intin_order[maxv], post_order[maxv], lch[maxv], rch[maxv];

int n;

 

boolread_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;

}

 

//把in_order[L1..R1]和post_order[L2..R2]建成一棵二叉树，返回树根

int build(int L1,int R1, int L2, int R2)

{

    if(L1 > R1)

    {

        return 0;//空树

    }

    int root = post_order[R2];

    int p = L1;

    while(in_order[p] != root)

    {

        p++;

    }

    int cnt = p-L1;//左子树的结点个数

    lch[root] = build(L1, p-1, L2, L2+cnt-1);

    rch[root] = build(p+1, R1, L2+cnt, R2-1);

    return root;

}

 

int best,best_sum; //目前为止的最优解和对应的权和

 

void dfs(int u,int sum)

{

    sum += u;

    if(!lch[u] && !rch[u])//叶子

    {

        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))

    {

        read_list(post_order);

        build(0, n-1, 0, n-1);

        best_sum = 1000000000;

        dfs(post_order[n-1], 0);

        cout << best <<"\n";

    }

    return 0;

}