CCCC L2-006 树的遍历

链接：https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456

题解:根据后序中序建树，层序遍历输出的裸题，建树的基本思想是利用递归，层序遍历可以利用队列，详细参见代码

代码：

 1 #include<bits/stdc++.h>
 2 #define inf 0x3f3f3f3f
 3 using namespace std;
 4 typedef long long ll;
 5 const int M = int(1e5) * 2 + 5;
 6 int in[M], po[M];
 7 int ptr;
 8 struct node {
 9     int l, r, v;
10 }t[M];
11 void build(int l, int r, int n)
12 {
13     if (l >= r)
14     {
15         t[n].v = inf;
16         return;
17     }
18     int root = po[ptr--];
19     t[n].v = root;
20     t[n].l = n * 2;
21     t[n].r = n * 2 + 1;
22     int mid = find(in, in + r, root) - in;
23     build(mid + 1, r, n * 2 + 1);
24     build(l, mid, n * 2);
25 }
26 void Print()
27 {
28     queue<int> q;
29     q.push(1);
30     int id;
31     while (!q.empty())
32     {
33         id = q.front();
34         q.pop();
35         if (t[id].v != inf)
36         {
37             if (id != 1)
38                 cout << " ";
39             cout << t[id].v;
40             q.push(t[id].l);
41             q.push(t[id].r);
42         }
43     }
44 }
45 int main()
46 {
47     int n; cin >> n;
48     for (int i = 0; i < n; i++) cin >> po[i];
49     for (int i = 0; i < n; i++) cin >> in[i];
50     ptr = n - 1;
51     build(0, n, 1);
52     Print();
53     return 0;
54 }