<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) {}
 * };
 */
class Solution {
public:
    TreeNode *constructFromPrePost(vector<int> &preorder, vector<int> &postorder) {
        if (preorder.empty()) { // 空节点
            return nullptr;
        }
        if (preorder.size() == 1) { // 叶子节点
            return new TreeNode(preorder[0]);
        }
        int left_size = ranges::find(postorder, preorder[1]) - postorder.begin() + 1; // 左子树的大小
        vector<int> pre1(preorder.begin() + 1, preorder.begin() + 1 + left_size);
        vector<int> pre2(preorder.begin() + 1 + left_size, preorder.end());
        vector<int> post1(postorder.begin(), postorder.begin() + left_size);
        vector<int> post2(postorder.begin() + left_size, postorder.end() - 1);
        TreeNode *left = constructFromPrePost(pre1, post1);
        TreeNode *right = constructFromPrePost(pre2, post2);
        return new TreeNode(preorder[0], left, right);
    }
};
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