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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func constructFromPrePost(preorder []int, postorder []int) *TreeNode {
	pos := map[int]int{}
	for i, x := range postorder {
		pos[x] = i
	}
	var dfs func(int, int, int, int) *TreeNode
	dfs = func(a, b, c, d int) *TreeNode {
		if a > b {
			return nil
		}
		root := &TreeNode{Val: preorder[a]}
		if a == b {
			return root
		}
		i := pos[preorder[a+1]]
		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, len(preorder)-1, 0, len(postorder)-1)
}