PAT-1020 Tree Traversals （25 分）【建树+bfs】

1020 Tree Traversals （25 分）
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

题意：
依次给出后序和中序遍历结果，输出中序的遍历结果
思路：
传统的建树，然后BFS遍历

#include
using namespace std;

const int maxn=50;
int pre[maxn],in[maxn],post[maxn];
int n;
struct Node
{
    int data;
    Node* lchild;
    Node* rchild;
};
Node* creat(int postL,int postR,int inL,int inR)
{
    if(postL>postR)
    {
        return NULL;
    }
    Node* root=new Node;
    root->data=post[postR];
    int k;
    for( k=inL;k<=inR;k++)
    {
        if(in[k]==post[postR])
            break;
    }
    int numLeft=k-inL;
    root->lchild=creat(postL,postL+numLeft-1,inL,k-1);
    root->rchild=creat(postL+numLeft,postR-1,k+1,inR);
    return root;
}
int num=0;

void BFS(Node* root){
    queue q;
    q.push(root);
    while (!q.empty())
    {
        Node* now = q.front();
        q.pop();
        cout<data;
        num++;
        if (numlchild!=NULL)
        {
            q.push(now->lchild);
        }
        if (now->rchild!=NULL)
        {
            q.push(now->rchild);
        }
    }
}

int main()
{
    scanf("%d",&n);
    for(int i=0;i