Solution
Approach #1 Recursive Solution[Accepted]
The current solution is very simple. We make use of a function construct(nums, l, r)
, which returns the maximum binary tree consisting of numbers within the indices and in the given array(excluding the element).
The algorithm consists of the following steps:

Start with the function call construct(nums, 0, n)
. Here, refers to the number of elements in the given array.

Find the index, , of the largest element in the current range of indices . Make this largest element, $ as the local root node.

Determine the left child using construct(nums, l, max_i)
. Doing this recursively finds the largest element in the subarray left to the current largest element.

Similarly, determine the right child using construct(nums, max_i + 1, r)
.

Return the root node to the calling function.
Java
public class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
return construct(nums, 0, nums.length);
}
public TreeNode construct(int[] nums, int l, int r) {
if (l == r)
return null;
int max_i = max(nums, l, r);
TreeNode root = new TreeNode(nums[max_i]);
root.left = construct(nums, l, max_i);
root.right = construct(nums, max_i + 1, r);
return root;
}
public int max(int[] nums, int l, int r) {
int max_i = l;
for (int i = l; i < r; i++) {
if (nums[max_i] < nums[i])
max_i = i;
}
return max_i;
}
}
Complexity Analysis

Time complexity : . The function construct
is called times. At each level of the recursive tree, we traverse over all the elements to find the maximum element. In the average case, there will be a levels leading to a complexity of . In the worst case, the depth of the recursive tree can grow upto , which happens in the case of a sorted array, giving a complexity of .

Space complexity : . The size of the can grow upto in the worst case. In the average case, the size will be for elements in , giving an average case complexity of
Analysis written by: @vinod23