PAT甲级1020（算法笔记9.2.4）
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 (≤), 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<queue>
using namespace std;
const int maxn = 31;
int post[maxn], in[maxn];
struct node
{
int data;
node* lchid;
node* rchid;
};
int n;
node* creat(int pl, int pr, int il, int ir)//中序后序建立二叉树
{
if (pl > pr)return NULL;
node* root=new node;
root->data= post[pr];
int k;
for ( k =il; k<=ir; k++)
if (in[k] == post[pr])
break;
int numl = k - il;
root->lchid = creat( pl, pl + numl - 1, il, k - 1);
root->rchid = creat(pl + numl, pr - 1, k + 1, ir);
return root;
}
int num=0;
void BFS(node*root)//层次遍历的广度优先搜索法
{
queue<node*>q;
q.push(root);
while (!q.empty())
{
node* T= q.front();
q.pop();
cout <<T->data;
num++;
if (num != n)
cout << " ";
if (T->lchid != NULL)
q.push(T->lchid);
if (T->rchid != NULL)
q.push(T->rchid);
}
cout << endl;
}
int main()
{
FILE* stream1;
freopen_s(&stream1, "input.txt", "r", stdin);
cin >> n;
for (int i = 1; i <= n; i++)
cin >> post[i];
for (int i = 1; i <= n; i++)
cin >>in[i];
node* root = creat(1, n, 1, n);
BFS(root);
return 0;
}