<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 {
    unordered_set<int> visited;
    TreeNode* rebuild(vector<int> pre, vector<int> in, int n, int s, int e){
        if(n > pre.size()){
            return NULL;
        }
        if(s > e){
            return NULL;
        }
        int start = pre[n];
        visited.insert(start);
        TreeNode* root = new TreeNode(start);
        int i = 0;
        for(; i<in.size(); i++){
            if(start == in[i])break;
        }
        if(i>0)
            root->left = rebuild(pre,in,n+1, s, i-1);
        while(n < pre.size() && visited.find(pre[n])!=visited.end()){
            n++;
        }
        if(i<in.size()-1 && visited.size()<pre.size())
            root->right = rebuild(pre,in,n,i+1,e);
        return root;
    }
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* p = rebuild(preorder,inorder,0,0,preorder.size());
        return p;
    }
};

static int _performance_boost_ = []()
{
    ios_base::sync_with_stdio(false),
    cin.tie(0),
    cout.tie(0);
    return 0;
} ();

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