Construct Binary Tree from Preorder and Postorder Traversal
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0889.Construct%20Binary%20Tree%20from%20Preorder%20and%20Postorder%20Traversal/images/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>
Tree; Array; Hash Table; Divide and Conquer; Binary Tree
/**
 * 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) {
        unordered_map<int, int> pos;
        int n = postorder.size();
        for (int i = 0; i < n; ++i) {
            pos[postorder[i]] = i;
        }
        function<TreeNode*(int, int, int, int)> dfs = [&](int a, int b, int c, int d) -> TreeNode* {
            if (a > b) {
                return nullptr;
            }
            TreeNode* root = new TreeNode(preorder[a]);
            if (a == b) {
                return root;
            }
            int i = pos[preorder[a + 1]];
            int m = i - c + 1;
            root->left = dfs(a + 1, a + m, c, i);
            root->right = dfs(a + m + 1, b, i + 1, d - 1);
            return root;
        };
        return dfs(0, n - 1, 0, n - 1);
    }
};