Maximum Binary Tree
<p>You are given an integer array <code>nums</code> with no duplicates. A <strong>maximum binary tree</strong> can be built recursively from <code>nums</code> using the following algorithm:</p>

<ol>
	<li>Create a root node whose value is the maximum value in <code>nums</code>.</li>
	<li>Recursively build the left subtree on the <strong>subarray prefix</strong> to the <strong>left</strong> of the maximum value.</li>
	<li>Recursively build the right subtree on the <strong>subarray suffix</strong> to the <strong>right</strong> of the maximum value.</li>
</ol>

<p>Return <em>the <strong>maximum binary tree</strong> built from </em><code>nums</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/0600-0699/0654.Maximum%20Binary%20Tree/images/tree1.jpg" style="width: 302px; height: 421px;" />
<pre>
<strong>Input:</strong> nums = [3,2,1,6,0,5]
<strong>Output:</strong> [6,3,5,null,2,0,null,null,1]
<strong>Explanation:</strong> The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
    - The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
        - Empty array, so no child.
        - The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
            - Empty array, so no child.
            - Only one element, so child is a node with value 1.
    - The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
        - Only one element, so child is a node with value 0.
        - Empty array, so no child.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0654.Maximum%20Binary%20Tree/images/tree2.jpg" style="width: 182px; height: 301px;" />
<pre>
<strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> [3,null,2,null,1]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
	<li>All integers in <code>nums</code> are <strong>unique</strong>.</li>
</ul>
Stack; Tree; Array; Divide and Conquer; Binary Tree; Monotonic Stack
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn construct(nums: &Vec<i32>, start: usize, end: usize) -> Option<Rc<RefCell<TreeNode>>> {
        if start >= end {
            return None;
        }
        let mut idx = 0;
        let mut max_val = -1;
        for i in start..end {
            if nums[i] > max_val {
                idx = i;
                max_val = nums[i];
            }
        }
        Some(Rc::new(RefCell::new(TreeNode {
            val: max_val,
            left: Self::construct(nums, start, idx),
            right: Self::construct(nums, idx + 1, end),
        })))
    }

    pub fn construct_maximum_binary_tree(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::construct(&nums, 0, nums.len())
    }
}
