<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 (postorder.size() == 1) {
            /* Only 1 element */
            TreeNode *a = new TreeNode(postorder[0]);
            return a;
        }
        /* The base element will always be the last one traversed in postorder */
        int base = postorder[postorder.size() - 1];
        int i = 0;
        while (inorder[i] != base) {
            i++;
        }
        /* The elements on the left side are those before the base, 
        and the elements on the right side are those after the base. */
        vector<int> newInOrder(inorder);
        vector<int> newPostOrder(postorder);
        if (i == 0) {
            /* There are only elements on the right side */
            newInOrder.erase(newInOrder.begin());
            newPostOrder.erase(newPostOrder.end());
            TreeNode *right = new TreeNode(base, nullptr, buildTree(newInOrder, newPostOrder));
            return right;
        }
        else if (i == inorder.size() - 1) {
            /* There are only elements on the left side */
            newInOrder.erase(newInOrder.end());
            newPostOrder.erase(newPostOrder.end());
            TreeNode *left = new TreeNode(base, buildTree(newInOrder, newPostOrder), nullptr);
            return left;
        }
        else {
            /* We'll need to split it up into left and right sides */
            vector<int> leftInOrder(inorder);
            vector<int> rightInOrder(inorder);
            vector<int> leftPostOrder(postorder);
            vector<int> rightPostOrder(postorder);
            leftInOrder.erase(leftInOrder.begin() + i, leftInOrder.end());
            rightInOrder.erase(rightInOrder.begin(), rightInOrder.begin() + i + 1);
            // int firstOnRight = inorder[i+1];
            // int j = 0;
            // while (postorder[j] != firstOnRight) {
            //     j++;
            // }
            leftPostOrder.erase(leftPostOrder.begin() + i, leftPostOrder.end());
            rightPostOrder.erase(rightPostOrder.begin(), rightPostOrder.begin() + i);
            rightPostOrder.erase(rightPostOrder.end());
            TreeNode *out = new TreeNode(base, 
                                        buildTree(leftInOrder, leftPostOrder), 
                                        buildTree(rightInOrder, rightPostOrder)
                                        );
            return out;
        }
    }
};
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