You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:


	Create a root node whose value is the maximum value in nums.
	Recursively build the left subtree on the subarray prefix to the left of the maximum value.
	Recursively build the right subtree on the subarray suffix to the right of the maximum value.


Return the maximum binary tree built from nums.

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Example 1:

Input: nums = [3,2,1,6,0,5]
Output: [6,3,5,null,2,0,null,null,1]
Explanation: 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.


Example 2:

Input: nums = [3,2,1]
Output: [3,null,2,null,1]


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Constraints:


	1 &lt;= nums.length &lt;= 1000
	0 &lt;= nums[i] &lt;= 1000
	All integers in nums are unique.


class Solution:
	def constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:
		if not nums:
			return None
		m = max(nums)
		idx  = nums.index(m)
		root = TreeNode(m)
		root.left = self.constructMaximumBinaryTree(nums[:idx])
		root.right = self.constructMaximumBinaryTree(nums[idx+1:])
		return root
class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        TreeNode root = construct(nums,0,nums.length-1);
        return root;
    }
    
    private TreeNode construct(int []nums, int start, int end) {
        if(start > end)
            return null;
        if(start == end)
            return new TreeNode(nums[start]);
        
        int maxIdx = findMax(nums,start,end);
        TreeNode root = new TreeNode(nums[maxIdx]);
        
        root.left = construct(nums,start,maxIdx-1);
        root.right = construct(nums,maxIdx+1,end);
        
        return root;
    }
    
    private int findMax(int []arr, int low, int high) {
        int idx = -1, max = Integer.MIN_VALUE;
        for(int i=low;i<=high;++i) 
            if(arr[i]>max) {
                max = arr[i];
                idx = i;
            }
        return idx;
    }
}
/**
 * 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* constructMaximumBinaryTree(vector<int>& nums) 
    {
        return recur(0, nums.size() - 1, nums);
    }
    
    TreeNode* recur(int st, int en, vector<int> &ar)
    {
        if(st > en)
        {
            return NULL;
        }
        TreeNode* cur = new TreeNode();
        int maxm = -1, ind;
        for(int i = st; i <= en; i++)
        {
            if(maxm < ar[i])
            {
                maxm = ar[i];
                ind = i;
            }
        }
        
        cur->val = ar[ind];
        cur->left = recur(st, ind-1, ar);
        cur->right = recur(ind+1, en, ar);
        
        return cur;
    }
};
var constructMaximumBinaryTree = function(nums) {
    if (nums.length === 0) return null;
    let max = Math.max(...nums);
    let index = nums.indexOf(max);
    let left = nums.slice(0, index);
    let right = nums.slice(index + 1);
    let root = new TreeNode(max);
    root.left = constructMaximumBinaryTree(left);
    root.right = constructMaximumBinaryTree(right);
    return root;
};
Maximum Binary Tree