PAT 甲级 1020 Tree Traversals

1020 Tree Traversals （25 point(s)）

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order 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 (≤30), the total number of nodes in the binary tree. The second line gives the postorder 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 level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7      

Sample Output:

4 1 6 3 5 7 2      

Experiential Summing-up

Accepted Code

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
const int INF=0x3fffffff;
const int maxn=40;
int n,post[maxn],in[maxn];
vector<int> ans;
struct node
{
  node * lchild, * rchild;
  int data;
};
node * create(int inL,int inR,int postL,int postR)
{
  if(postL>postR)
    return NULL;
  int k;
  for(k=inL;k<=inR;++k)
  {
    if(in[k]==post[postR])
    {
      break;
    }
  }
  node * a=new node;
  a->data=post[postR];
  a->lchild=create(inL,k-1,postL,postL+(k-inL)-1);
  a->rchild=create(k+1,inR,postL+(k-inL),postR-1);
  return a;
}
void level(node * root)
{
  queue<node *> q;
  q.push(root);
  while(q.size())
  {
    node * x=q.front();
    ans.push_back(x->data);
    q.pop();
    if(x->lchild!=NULL)
      q.push(x->lchild);
    if(x->rchild!=NULL)
      q.push(x->rchild);
  }
}
int main()
{
  scanf("%d",&n);
  for(int i=0;i<n;++i)
    scanf("%d",&post[i]);
  for(int i=0;i<n;++i)
    scanf("%d",&in[i]);
  node * root=create(0,n-1,0,n-1);
  level(root);
  for(int i=0;i<ans.size();++i)
  {
    printf("%d",ans[i]);
    if(i!=ans.size()-1)
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}