知道后续遍历和中序遍历还原二叉树并求层析遍历

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题目：

输入n代表二叉树节点数量，接下来两行分边n个整数，代表二叉树的后序遍历和中序遍历

输出层次遍历的顺序

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <fstream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <iomanip>using namespace std;//#pragma comment(linker, "/STACK:102400000,102400000")#define maxn 10050#define MOD 1000000007#define mem(a , b) memset(a , b , sizeof(a))#define LL long long#define ULL unsigned long longtypedef pair<int , int> pii;const long long INF= 0x3fffffff;int last[maxn] , mid[maxn];int a[maxn] , b[maxn];struct node{    int l , r , val;}tmp , ans[maxn];int res[maxn] , id;void solve1(int root , int l1 , int r1 , int l2 , int r2);   //给定左子树的区间void solve2(int root , int l1 , int r1 , int l2 , int r2)    //给定右子树的期间{    //cout << root << endl;    if(r1 < l1 || l1 > r1 || r2 < l2 || l2 > r2)    {        ans[root].r = -1;        return;    }    if(l1 == r1)    {        ans[root].r = a[r1];        return ;    }    ans[root].r = a[r1];    solve1(ans[root].r , l1 , l1 + mid[a[r1]] - l2 - 1 , l2 , mid[a[r1]] - 1 );    solve2(ans[root].r , l1 + mid[a[r1]] - l2  , r1 - 1 , mid[a[r1]]+1 , r2);}void solve1(int root , int l1 , int r1 , int l2 , int r2){    //cout << root << endl;    if(r1 < l1 || l1 > r1 || r2 < l2 || l2 > r2)    {        ans[root].l = -1;        return;    }    if(l1 == r1)    {        ans[root].l = a[r1];        return ;    }    ans[root].l = a[r1];    solve1(ans[root].l , l1 , l1 + mid[a[r1]] - l2 - 1 , l2 , mid[a[r1]] - 1 );    solve2(ans[root].l , l1 + mid[a[r1]] - l2  , r1-1 , mid[a[r1]] +1 , r2);}void BFS(int root){    queue<int>q;    while(!q.empty()) q.pop();    q.push(root);    while(!q.empty())    {        int val = q.front();        q.pop();        res[id++] = val;        if(ans[val].l != -1) q.push(ans[val].l);        if(ans[val].r != -1) q.push(ans[val].r);       // cout << val << " " << ans[val].l << " " << ans[val].r << endl;    }}int main(){    int n , m , num , l , r;    while(scanf("%d" , &n) != EOF)    {        id = 0;        mem(last , 0);        mem(mid , 0);        mem(ans , -1);        for(int i = 1 ; i <= n ; i ++)        {            scanf("%d" , &a[i]);            last[a[i]] = i;        }        for(int i = 1 ; i <= n ; i ++)        {            scanf("%d" , &b[i]);            mid[b[i]] = i;        }        int root = a[n];        solve1(a[n] , 1 , mid[root]-1 , 1 , mid[root] - 1);        solve2(a[n] , mid[root] , n-1 , mid[root] + 1 , n);       // cout << "YES" << endl;        BFS(root);        for(int i = 0 ; i < id ; i ++)        {            printf("%d" , res[i]);            if(i != id -1) printf(" ");        }        cout << endl;    }    return 0;}

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