已知二叉树的先序遍历和中序遍历，输出它的后序遍历

---

 

#include<iostream>
#include<vector>
#include<string>
#include<unordered_map>
using namespace std;

unordered_map<int, int>record;
vector<char>res;
//先序遍历数组，中序遍历数组，当前处理的先序遍历数组的开始索引，当前处理的先序遍历数组的结束索引，当前处理的中序遍历数组的开始索引，当前处理的中序遍历数组的结束索引，
//先序遍历数组，中序遍历数组， 
void my_buildTree(vector<char>& preorder, vector<char>& inorder, int pre_left, int pre_right, int in_left, int in_right)
{
  //当前遍历的先序数组，元素已经没有了，返回空
  if (pre_left>pre_right)return;
  // 在中序遍历中定位根节点
  int root_index = record[preorder[pre_left]];

  //创建一个根节点
  

  //左子树的个数
  int left_tree = root_index - in_left;
  //构建左子树
  my_buildTree(preorder, inorder,
    pre_left + 1,
    pre_left + left_tree,
    in_left,
    root_index - 1);
  //构建右子树
  my_buildTree(preorder, inorder,
    pre_left + left_tree + 1,
    pre_right,
    root_index + 1,
    in_right);
  res.push_back(inorder[root_index]);
}
int main()
{
  string pre;
  string in;
  cin >> pre >> in;
  vector<char> preorder;
  vector<char> inorder;
  for (int i = 0; i<pre.size(); i++)
  {
    preorder.push_back(pre[i]);
    inorder.push_back(in[i]);
  }
  int n = inorder.size();
  
  //res.resize(n);
  for (int i = 0; i < n; i++)
  {
    record[inorder[i]] = i;
  }
  my_buildTree(preorder, inorder, 0, n - 1, 0, n - 1);

  for (int i = 0; i<res.size(); i++)
    cout << res[i]<<" ";
  cout << endl;

}