PAT 1020 Tree Traversals(二叉の木の中の順序に従って、順番に階層を巡回する)

PAT 1020
1020 Tree Traversals (25分)
Suppose that all the keys in a binary tree are distinct positive integers.Gven the postorder and in order trversal sequences,you are supposed to out put the level order trure sal sequence of the corestinder.
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 e e e separated by space.
Output Specification:
For each test case,print in one line the level order trversal sequence of the cores ponding binary tree.All the numbers in a line must be separated by exactlyone space,and the must beのextra space space space the the the the the end of line.line.ine.ine.line.ine.ine.ine.line.ine.ine
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

配列の書き込みがこすられています.各層のノード数が小さいと、層数が大きいため、配列が貯蓄できなくなります.
チェーンを変えてみましょう.針が本当に痛いです. 
どのようにして既知の遍歴によって二叉樹を復元しますか?
テンプレートコードがあります.
詳しくはここに移す

#include
#include
#include
#include
using namespace std;
typedef struct Bitree {
    int data;
    struct Bitree *left;
    struct Bitree *right;
} tree,*Tree;
tree *Resume( int in[],int post[], int len) {
    if(len<=0)
        return NULL;
    tree* temp=new tree;
    temp->data=*post;
    int i=0;
    while(idata)
            break;
        i++;
    }
    temp->left=Resume(in,post-len+i,i);
    temp->right=Resume(in+i+1,post-1,len-i-1);
    return temp;
}
queueq;
int res[100000+50];
int len=0;
void dfs(Tree t) {
    q.push(t);
    while(!q.empty()) {
        t=q.front();
        q.pop();
        res[len++]=t->data;
        if(t->left!=NULL)
            q.push(t->left);
        if(t->right!=NULL)
            q.push(t->right);
    }
}
int in[100],post[100];
int main() {
    int n;
    cin>>n;
    for(int i=0;i