PAT甲級 1020 Tree Traversals（二叉樹的遍歷與建樹）

#include <iostream>
#include<cstdio>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;

const int maxn = 30;
int n;
int postOrd[maxn + 5], inOrd[maxn + 5];
vector<int> layer[30];

struct node{
    int data;
    node* left;
    node* right;
};
//由後序遍歷和中序遍歷建樹
node* create(int postL, int postR, int inL, int inR, int deep){
    //邊界條件，不存在該後序遍歷序列，退出
    if(postL > postR) return NULL;
    //創建新節點並賦值
    node* root = new node;
    root->data = postOrd[postR];//後序遍歷最後一個數是根節點
    //找到中序序列中該根節點的下標並保存在k中
    layer[deep].push_back(postOrd[postR]);

    int k;
    for(int i = 0; i < n; i++){
        if(inOrd[i] == postOrd[postR]){
            k  = i;
            break;
        }
    }
    //左子樹節點個數爲k-inL,分別求該節點的左兒子和右兒子..
    int numLeft = k - inL;
    root->left = create(postL, postL+numLeft - 1, inL, k-1, deep+1);
    root->right = create(postL+numLeft, postR-1, k+1, inR, deep+1);

    return root;
}

int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &postOrd[i]);
    }
    for(int i = 0; i < n; i++){
        scanf("%d", &inOrd[i]);
    }
    node* root = create(0, n-1, 0, n-1, 1);
    int flag = 0;
    for(int i = 1; i <30; i++){
        int len = layer[i].size();
        for(int j = 0; j < len; j++){
            if(flag) printf(" ");
            printf("%d", layer[i][j]);
            flag = 1;
        }
    }

    return 0;
}