Tree UVA - 548 已知中序遍历和后序遍历，求这颗二叉树。

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

/**
题目：Tree UVA - 548
链接：https://vjudge.net/problem/UVA-548
题意：算法竞赛入门经典P155 eg6-8
思路：后序遍历的最后一个为根。那么知道了根是多少，就可以在中序遍历找到根的位置，
根的左边为左子树，右边为右子树。左子树，右子树的后序遍历也可以通过原来的后序遍历中分成两部分获得。
后序遍历和中序遍历长度相同。
递归处理每一个结点即可。

收获：
char s[];
stringstream ss(s);
n = 0;
int x;
while(ss>>x) a[n++] = x;

处理一行由数字和空格组成的字符串，划分成数字的方式。
*/

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> P;
const int maxn = 1e4+;
const int mod = 1e9+;
int a[maxn], b[maxn], n;
struct node
{
    int value;
    node *left, *right;
    node():left(NULL),right(NULL){}
};
node *root;
char s[*maxn];
void Input(char *s,int a[])
{
    stringstream ss(s);
    n = ;
    int x;
    while(ss>>x) a[n++] = x;
}
node* build(int l,int r,int p)
{
    if(l>r){
        return NULL;
    }
    node *x = new node();
    int mid;
    for(int i = l; i <= r; i++){
        if(a[i]==b[p]){
            mid = i; break;
        }
    }
    x->value = a[mid];
    x->left = build(l,mid-,p--(r-mid));///可推算出根的位置。
    x->right = build(mid+,r,p-);
    return x;
}
int ans, leaf;
void Find(node *root,int sum)
{
    if(root->left==NULL&&root->right==NULL){
        if(sum+root->value<ans){
            ans = sum + root->value;
            leaf = root->value;
        }else
        {
            if(sum+root->value==ans&&root->value<leaf){
                leaf = root->value;
            }
        }
    }else
    {
        if(root->left!=NULL){
            Find(root->left,sum+root->value);
        }
        if(root->right!=NULL){
            Find(root->right,sum+root->value);
        }
    }
}
int main()
{
    while(gets(s)!=NULL){
        Input(s,a);
        gets(s);
        Input(s,b);
        root = build(,n-,n-);///中序遍历[0,n-1]，第三个n-1表示根。
        ans = maxn*maxn;
        Find(root,);
        printf("%d\n",leaf);
    }
    return ;
}