654. Maximum Binary Tree

Problem

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.

  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.

Solution (Java)

/**
 * 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 {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return mbt(nums, 0, nums.length - 1);
    }

    private TreeNode mbt(int[] nums, int l, int r) {
        if (l > r || l >= nums.length || r < 0) {
            return null;
        }
        if (l == r) {
            return new TreeNode(nums[r]);
        }
        int max = Integer.MIN_VALUE;
        int maxidx = 0;
        for (int i = l; i <= r; i++) {
            if (nums[i] > max) {
                max = nums[i];
                maxidx = i;
            }
        }
        TreeNode root = new TreeNode(max);
        root.left = (mbt(nums, l, maxidx - 1));
        root.right = (mbt(nums, maxidx + 1, r));
        return root;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).