<p>Given two integer arrays <code>preorder</code> and <code>inorder</code> where <code>preorder</code> is the preorder traversal of a binary tree and <code>inorder</code> is the inorder traversal of the same tree, construct and return <em>the binary tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> preorder = [-1], inorder = [-1]
<strong>Output:</strong> [-1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 3000</code></li>
	<li><code>inorder.length == preorder.length</code></li>
	<li><code>-3000 &lt;= preorder[i], inorder[i] &lt;= 3000</code></li>
	<li><code>preorder</code> and <code>inorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>inorder</code> also appears in <code>preorder</code>.</li>
	<li><code>preorder</code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {

    vector<int> po,io;
    int n;
    bool b[6001];
    unordered_map<int,int> pos;

public:

    TreeNode* left(int l, int r){
        TreeNode *ret;
        ret=new TreeNode();
        if (l==r){
            ret->val=po[l];
            b[3000+po[l]]=1;
        }else{
            int k=0;
            int lim=pos[po[l]];
            for (int i=0;i<lim;i++) {
                if (!b[3000+io[i]]) k++;
            }
            b[3000+po[l]]=1;
            ret->val=po[l];
            if (k) ret->left=left(l+1,l+k);
            if (l+k+1<=r) ret->right=right(l+k+1,r);
        }
        return ret;
    }  
    TreeNode* right(int l, int r){
        TreeNode *ret;
        ret=new TreeNode();
        if (l==r){
            ret->val=po[l];
            b[3000+po[l]]=1;
        }else{
            int k=0;
            int lim=pos[po[l]];
            for (int i=0;i<lim;i++) {
                if (!b[3000+io[i]]) k++;
            }
            b[3000+po[l]]=1;
            ret->val=po[l];
            if (k) ret->left=left(l+1,l+k);
            if (l+k+1<=r) ret->right=right(l+k+1,r);
        }
        return ret;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
      n=preorder.size();
      for (int i=0;i<n;i++){
        pos[inorder[i]]=i;
        po.push_back(preorder[i]);
        io.push_back(inorder[i]);
      } 
      for (int i=0;i<6001;i++) b[i]=0;
      TreeNode *head;
      head=new TreeNode();
      head->val=preorder[0];
      b[3000+preorder[0]]=1;
      if (n<2) return head;
      int k=0;
      for (int i=0;i<pos[preorder[0]];i++){
        if (!b[3000+inorder[i]])k++;
      } 
      if (k!=0) head->left=left(1,k);
      if (k+1<=n-1) head->right=right(k+1,n-1);

      return head;
    }
};
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