浙大pat1020题解

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

32000 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 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

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
#define MaxSize 32

struct Node
{
int data;
Node *left;
Node *right;
};
queue<Node*>result;
vector<int> vec;
Node *CreatNode(int *post,int*in,int len)
{
if(len == NULL)
return NULL;
int i = len-1;
while(in[i]!=post[len-1]) i--;
Node *p;
p =new Node;
p->data = in[i];
p->left = CreatNode(post,in,i);
p->right = CreatNode(post+i,in+i+1,len-i-1);
return p;
}

void LevelPrint(Node*p)
{
if(p!=NULL) result.push(p);
while(!result.empty())
{
Node * temp = result.front();
if(temp->left!=NULL) result.push(temp->left);
if(temp->right!=NULL) result.push(temp->right);
vec.push_back(temp->data);
result.pop();
}
}

int main()
{
int Post[MaxSize],In[MaxSize];
int len;
cin >> len;
for(int i=0;i<len;i++)
cin >> Post[i];
for(int j=0;j<len;j++)
cin >> In[j];
Node*root = CreatNode(Post,In,len);
LevelPrint(root);
int i;
for(i=0;i<vec.size()-1;i++)
cout<<vec[i]<<" ";
cout<<vec[i]<<endl;
return 0;
}