Pre-inorder reconstruction binary tree

Topic description

Input the result of preorder traversal and inorder traversal of a binary tree , please rebuild the binary tree.

Notice:

The value of each node in the binary tree is different from each other;
the input pre-order traversal and in-order traversal must be legal;

Data range
The number of nodes in the tree ranges from [0,100].

Example
Given:
Preorder traversal is: [3, 9, 20, 15, 7]
Inorder traversal is: [9, 3, 15, 20, 7]

Returns: [3, 9, 20, null, null, 15, 7, null, null, null, null]
The returned binary tree looks like this:

    3
   / \
  9  20
    /  \
   15   7

CPP

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    
    
public:

    unordered_map<int, int> hash;
    
    TreeNode* dfs(vector<int>& pre, vector<int>& in, int pl, int pr, int il, int ir) {
    
    
        if (pl > pr) return NULL;
        // 用中序遍历确定，根结点左子树的大小
        int k = hash[pre[pl]] - il;
        TreeNode* root = new TreeNode(pre[pl]);
        // 边界可以举一个实际的例子来确定
        root->left = dfs(pre, in, pl + 1, pl + k, il, il + k - 1);
        root->right = dfs(pre, in, pl + k + 1, pr, il + k + 1, ir);
        return root;
    }
    
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
    
    
        int n = inorder.size();
        
        for (int i = 0; i < n; i++) hash[inorder[i]] = i;
        
        return dfs(preorder, inorder, 0, n-1, 0, n-1);
    }
};