<p>Given two integer arrays <code>inorder</code> and <code>postorder</code> where <code>inorder</code> is the inorder traversal of a binary tree and <code>postorder</code> is the postorder 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> inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= inorder.length &lt;= 3000</code></li>
	<li><code>postorder.length == inorder.length</code></li>
	<li><code>-3000 &lt;= inorder[i], postorder[i] &lt;= 3000</code></li>
	<li><code>inorder</code> and <code>postorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>postorder</code> also appears in <code>inorder</code>.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
	<li><code>postorder</code> is <strong>guaranteed</strong> to be the postorder 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 {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0){
            return nullptr;
        }
        if (inorder.size() == 1){
            return new TreeNode(inorder[0]);
        }
        int n = postorder.size();
        TreeNode* root = new TreeNode(postorder[n-1]);
        int i = 0;
        while (inorder[i] != root->val){
            i++;
        }
        vector<int> in_left(inorder.begin(), inorder.begin()+i);
        // vector<int> in_right;
        // vector<int> post_left;
        // vector<int> post_right;

        // vector<int> in_left;
        vector<int> in_right;
        vector<int> post_left;
        vector<int> post_right;
        // for (int j = 0; j < i; j++){
        //     in_left.push_back(inorder[j]);
        // }
        for (int j = i+1; j < n; j++){
            in_right.push_back(inorder[j]);
        }
        int l_size = in_left.size();
        for (int j = 0; j < l_size; j++){
            post_left.push_back(postorder[j]);
        }
        for (int j = l_size; j < n-1; j++){
            post_right.push_back(postorder[j]);
        }
        root->left = buildTree(in_left, post_left);
        root->right = buildTree(in_right, post_right);
        return root;
        
    }
};


// class Solution:
//     def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
//         n = len(postorder)
//         if n == 1:
//             return TreeNode(postorder[0])
//         def r(postord, inord):
//             if not postord: return None
//             if len(postord) == 1: return TreeNode(postord[0])
//             v = postord[-1]
//             root = TreeNode(v)
//             i = inord.index(v)
//             postl = postord[:i]
//             postr = postord[i:-1]
//             #prel = preord[1:i+1]
//             #prer = preord[i+1:]
//             inl = inord[:i]
//             inr = inord[i+1:]
//             root.left = r(postl, inl)
//             root.right = r(postr, inr)
//             return root
//         return r(postorder, inorder)        
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