根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

3

/ \
9 20
/ \
15 7

/**
 * 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:
    map<int, int> hash;
    vector<int> preorder1;

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        preorder1 = preorder;
        int n = inorder.size();
        for (int i = 0; i < n; i++)
            hash[inorder[i]] = i;
        return dfs(0, preorder.size() - 1, 0, inorder.size() - 1);
    }

    TreeNode *dfs(int prel, int prer, int inol, int inor) {
        if (prel > prer) return nullptr;

        auto root = new TreeNode(preorder1[prel]);
        int k = hash[root->val];

        root->left = dfs(prel + 1, prel + k - inol, inol, k - 1);
        root->right = dfs(prel + k - inol + 1, prer, k + 1, inor);
        return root;
    }
};