(pat)A1138. Postorder Traversal

Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree.

Sample Input:
7
1 2 3 4 5 6 7
2 3 1 5 4 7 6
Sample Output:
3
昨天临睡前，交了一发没有全过，突然想起没有考虑左子树为空，右子树存在的情况，今天起来交了一发，过了。还是算模拟题啦，代码也很短。

#include<iostream>
#include<cstdio>
#include<stack>

using namespace std;

int pre[50005];
int in[50005];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        scanf("%d",pre+i);
    for(int i=0;i<n;i++)
        scanf("%d",in+i);
        int l,r,L,R;
        l=L=0;
        r=R=n-1;
    while(l<r)
    {
        int mid;
        for(mid=L;mid<=R;mid++)
            if(in[mid]==pre[l])
                break;
        if(L<mid)
        {
            R=mid-1;
            r=mid-L+l;
            l++;
        }
        else
        {
            l=mid-L+l+1;
            L=mid+1;
        }
    }
    printf("%d",pre[l]);
    return 0;
}