<p>Given two integer arrays, <code>preorder</code> and <code>postorder</code> where <code>preorder</code> is the preorder traversal of a binary tree of <strong>distinct</strong> values and <code>postorder</code> is the postorder traversal of the same tree, reconstruct and return <em>the binary tree</em>.</p>

<p>If there exist multiple answers, you can <strong>return any</strong> of them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/24/lc-prepost.jpg" style="width: 304px; height: 265px;" />
<pre>
<strong>Input:</strong> preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
<strong>Output:</strong> [1,2,3,4,5,6,7]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 30</code></li>
	<li><code>1 &lt;= preorder[i] &lt;= preorder.length</code></li>
	<li>All the values of <code>preorder</code> are <strong>unique</strong>.</li>
	<li><code>postorder.length == preorder.length</code></li>
	<li><code>1 &lt;= postorder[i] &lt;= postorder.length</code></li>
	<li>All the values of <code>postorder</code> are <strong>unique</strong>.</li>
	<li>It is guaranteed that <code>preorder</code> and <code>postorder</code> are the preorder traversal and postorder traversal of the same binary 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) {}
 * };
 */

constexpr int MAX = 32;

int a[MAX], b[MAX];

TreeNode* DFS(int l1, int r1, int l2, int r2) {
  TreeNode* ret = new TreeNode(a[l1]);
  int n = r1 - l1;
  if (n == 1) {
    return ret;
  }
  int x = a[l1 + 1], y = b[r2 - 2], p = l1, q = l2;
  for (; p < r1 && a[p] != y; ++p);
  for (; q < r2 && b[q] != x; ++q);
  if (x == y) {
    ret->left = DFS(l1 + 1, r1, l2, r2 - 1);
  } else {
    ret->left = DFS(l1 + 1, p, l2, q + 1);
    ret->right = DFS(p, r1, q + 1, r2 - 1);
  }
  return ret;
}

class Solution {
public:
    TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder) {
      int n = preorder.size();
      copy(preorder.begin(), preorder.end(), a);
      copy(postorder.begin(), postorder.end(), b);
      return DFS(0, n, 0, n);
    }
};
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