UVA548

Description

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

Ouput

For each tree description you should output the value of the leaf node of a path of least value. In thecase of multiple paths of least value you should pick the one with the least value on the terminal nod

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

Explanation

题目给出二叉树的中序遍历和后序遍历，我们据此可以构造出该二叉树

后序遍历的末位即为根结点，因此只需在中序遍历中找到该结点，即可得到根结点左右子树的中序遍历和后序遍历，递归左右子树

找最优方法为由根结点出发递归遍历直到叶子结点，将该路径的权和与目前的最优解取最大值赋给最优解变量

Code

#include <cstdio>#include <string>#include <sstream>#include <algorithm>#include <iostream>using namespace std;//因为各个节点的权值各不相同且都为整数，直接用权值作为节点编号const int maxv = 10000 + 10;int in_order[maxv],post_order[maxv],lch[maxv],rch[maxv];int n;bool read_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;}//根据inorder[L1...R1]和post[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;    int root = post_order[R2];    int p = L1;    while(in_order[p] != root)        p++;    int cnt = p-L1; //左子树的节点个数    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]){ //叶子节点        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;}