PAT 1138 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 (≤ 50,000), 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.

题解:
题中给定先序序列和中序序列后，需要输出后序序列第一个数字，其实和以前遇到的前中序转后序的题应该一致，就是在post（左子树），post（左子树），输出值的语句后面加上一个return就可以了，这样的话输出第一个值就会跳出函数。
代码：

#include<iostream>
#include<vector>
using namespace std;
//vector<int> pre, in;
int pre[50009],in[50009];
vector<int> post;
void postorder(int preL,int preR,int inL,int inR){
	if(preL>preR||inL>inR) return;
	int k=inL;
	while(in[k]!=pre[preL]) k++;
	int numL=k-inL;
	postorder(preL+1,preL+numL,inL,inL+numL-1);//左子树 
	postorder(preL+numL+1,preR,inL+numL+1,inR);//右子树 
	post.push_back(pre[preL]);
}
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>pre[i];
	for(int i=1;i<=n;i++) cin>>in[i];
	postorder(1,n,1,n);
	cout<<post[0];
	return 0;
}

题中遇到的问题：

一定要看清楚题目，测试的时候结果出现段错误，刚开始以为数组越界，后来发现是把50000看成了5000.