# 654. Maximum Binary Tree

## Description

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

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

Return the maximum binary tree built from nums.

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]


Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000
• All integers in nums are unique.

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int[] nums;

public TreeNode constructMaximumBinaryTree(int[] nums) {
this.nums = nums;
return dfs(0, nums.length - 1);
}

private TreeNode dfs(int l, int r) {
if (l > r) {
return null;
}
int i = l;
for (int j = l; j <= r; ++j) {
if (nums[i] < nums[j]) {
i = j;
}
}
TreeNode root = new TreeNode(nums[i]);
root.left = dfs(l, i - 1);
root.right = dfs(i + 1, r);
return root;
}
}

• /**
* 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 dfs(nums, 0, nums.size() - 1);
}

TreeNode* dfs(vector<int>& nums, int l, int r) {
if (l > r) return nullptr;
int i = l;
for (int j = l; j <= r; ++j) {
if (nums[i] < nums[j]) {
i = j;
}
}
TreeNode* root = new TreeNode(nums[i]);
root->left = dfs(nums, l, i - 1);
root->right = dfs(nums, i + 1, r);
return root;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:
def dfs(nums):
if not nums:
return None
val = max(nums)
i = nums.index(val)
root = TreeNode(val)
root.left = dfs(nums[:i])
root.right = dfs(nums[i + 1 :])
return root

return dfs(nums)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func constructMaximumBinaryTree(nums []int) *TreeNode {
var dfs func(l, r int) *TreeNode
dfs = func(l, r int) *TreeNode {
if l > r {
return nil
}
i := l
for j := l; j <= r; j++ {
if nums[i] < nums[j] {
i = j
}
}
root := &TreeNode{Val: nums[i]}
root.Left = dfs(l, i-1)
root.Right = dfs(i+1, r)
return root
}
return dfs(0, len(nums)-1)
}

• /**
* 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 constructMaximumBinaryTree(nums: number[]): TreeNode | null {
const n = nums.length;
if (n === 0) {
return null;
}
const [val, i] = nums.reduce((r, v, i) => (r[0] < v ? [v, i] : r), [-1, 0]);
return new TreeNode(
val,
constructMaximumBinaryTree(nums.slice(0, i)),
constructMaximumBinaryTree(nums.slice(i + 1)),
);
}


• // 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::rc::Rc;
use std::cell::RefCell;
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())
}
}