LeetCode in Kotlin

654. Maximum Binary Tree

Medium

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.

Solution

import com_github_leetcode.TreeNode

/*
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun constructMaximumBinaryTree(nums: IntArray): TreeNode? {
        return mbt(nums, 0, nums.size - 1)
    }

    private fun mbt(nums: IntArray, l: Int, r: Int): TreeNode? {
        if (l > r || l >= nums.size || r < 0) {
            return null
        }
        if (l == r) {
            return TreeNode(nums[r])
        }
        var max = Int.MIN_VALUE
        var maxidx = 0
        for (i in l..r) {
            if (nums[i] > max) {
                max = nums[i]
                maxidx = i
            }
        }
        val root = TreeNode(max)
        root.left = mbt(nums, l, maxidx - 1)
        root.right = mbt(nums, maxidx + 1, r)
        return root
    }
}

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