1138. Postorder Traversal (25) 二叉树前序中须转后序

1138. Postorder Traversal (25)

时间限制

600 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

71 2 3 4 5 6 72 3 1 5 4 7 6

Sample Output:

3

思路：万万没想到会考这么基础的题。。。给出二叉树的前序序列和中序序列求后序序列的第一的节点，都是基本操作

#include<cstdio>#include<vector>using namespace std;vector<int>pos;int n,pre[50000],ino[50000];void getpos(int ps,int pe,int is,int ie){if(ps>pe) return;if(ps==pe){pos.push_back(pre[pe]);}else{int i=is;while(i<ie&&ino[i]!=pre[ps]) i++;getpos(ps+1,ps+i-is,is,i-1);getpos(ps+i-is+1,pe,i+1,ie);pos.push_back(pre[ps]);}}int main(){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&pre[i]);}for(int i=0;i<n;i++){scanf("%d",&ino[i]);}getpos(0,n-1,0,n-1);printf("%d",pos[0]);return 0;}