题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856

注意点

这道题一个是镜像树的概念（其实就是反转二叉树，本来还想用后序遍历重写一个的），结果复制粘贴代码稍微修改一下就行了
另一个是插入函数需要引用，不然不会对原根结点进行改变

代码

  1. #include<algorithm>
  2. #include<vector>
  3. #include<cstdio>
  4. #include<iostream>
  5. using namespace std;
  7. int n;
  8. vector<int> pre(n), post(n), preM(n), postM(n), origin(n);
  9. struct Node{
  10.     int data;
  11.     Node *lchild;
  12.     Node *rchild;
  13. };
  15. void preOrder(Node *root){
  16.     if(root == NULL) return;
  17.     pre.push_back(root->data);
  18.     preOrder(root->lchild);
  19.     preOrder(root->rchild);
  20.     return;
  21. }
  23. void preOrderMirror(Node *root){
  24.     if(root == NULL) return;
  25.     preM.push_back(root->data);
  26.     preOrderMirror(root->rchild);
  27.     preOrderMirror(root->lchild);
  28.     return;
  29. }
  31. void postOrder(Node *root){
  32.     if(root == NULL) return;
  33.     postOrder(root->lchild);
  34.     postOrder(root->rchild);
  35.     post.push_back(root->data);
  36.     return;
  37. }
  39. void postOrderMirror(Node *root){
  40.     if(root == NULL) return;
  41.     postOrderMirror(root->rchild);
  42.     postOrderMirror(root->lchild);
  43.     postM.push_back(root->data);
  44.     return;
  45. }
  47. void insert(Node *&root,int num){
  48.     if(root == NULL){
  49.         root = new Node;
  50.         root->data = num;
  51.         root->lchild = NULL;
  52.         root->rchild = NULL;
  53.         return;
  54.     }
  55.     if(num < root->data) insert(root->lchild, num);
  56.     else insert(root->rchild, num);
  57.     return;
  58. }
  60. int main(){
  61.     Node *root = NULL;
  62.     int temp_num;
  63.     scanf("%d", &n);
  64.     for(int i = 0; i < n; i++){
  65.         scanf("%d",&temp_num);
  66.         origin.push_back(temp_num);
  67.         insert(root, temp_num);
  68.     }
  69.     preOrder(root);
  70.     preOrderMirror(root);
  71.     postOrder(root);
  72.     postOrderMirror(root);
  73.     if(origin == pre){
  74.         printf("YES\n");
  75.         for(int i = 0; i < post.size(); i++){
  76.             printf("%d",post[i]);
  77.             if(i != post.size() - 1) printf(" ");
  78.         }
  79.     } else if(origin == preM){
  80.         printf("YES\n");
  81.         for(int i = 0; i < postM.size(); i++){
  82.             printf("%d",postM[i]);
  83.             if(i != postM.size() - 1) printf(" ");
  84.         }
  85.     } else printf("NO\n");
  86.     return 0;
  87. }