Construct Binary Search Tree from Preorder Traversal
<p>Given an array of integers preorder, which represents the <strong>preorder traversal</strong> of a BST (i.e., <strong>binary search tree</strong>), construct the tree and return <em>its root</em>.</p>

<p>It is <strong>guaranteed</strong> that there is always possible to find a binary search tree with the given requirements for the given test cases.</p>

<p>A <strong>binary search tree</strong> is a binary tree where for every node, any descendant of <code>Node.left</code> has a value <strong>strictly less than</strong> <code>Node.val</code>, and any descendant of <code>Node.right</code> has a value <strong>strictly greater than</strong> <code>Node.val</code>.</p>

<p>A <strong>preorder traversal</strong> of a binary tree displays the value of the node first, then traverses <code>Node.left</code>, then traverses <code>Node.right</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1008.Construct%20Binary%20Search%20Tree%20from%20Preorder%20Traversal/images/1266.png" style="height: 386px; width: 590px;" />
<pre>
<strong>Input:</strong> preorder = [8,5,1,7,10,12]
<strong>Output:</strong> [8,5,10,1,7,null,12]
</pre>

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

<pre>
<strong>Input:</strong> preorder = [1,3]
<strong>Output:</strong> [1,null,3]
</pre>

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

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 100</code></li>
	<li><code>1 &lt;= preorder[i] &lt;= 1000</code></li>
	<li>All the values of <code>preorder</code> are <strong>unique</strong>.</li>
</ul>
Stack; Tree; Binary Search Tree; Array; Binary Tree; Monotonic Stack
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function bstFromPreorder(preorder: number[]): TreeNode | null {
    const dfs = (i: number, j: number): TreeNode | null => {
        if (i > j) {
            return null;
        }
        const root = new TreeNode(preorder[i]);
        let [l, r] = [i + 1, j + 1];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (preorder[mid] > preorder[i]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        root.left = dfs(i + 1, l - 1);
        root.right = dfs(l, j);
        return root;
    };
    return dfs(0, preorder.length - 1);
}
