654. Maximum Binary Tree

LeetCode 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.

Constraints:

• 1 <= nums.length <= 1000

• 0 <= nums[i] <= 1000

• All integers in nums are unique.

Tags

Tree

Solution

Recursion. The edge case is return null when the input array is empty. Find the maximum number in array as the value of the current node, and assign left max & right max to its left and right children respectively.

Complexity

• Time complexity: $O(n^2)$

• Space complexity: $O(n)$

Code

func constructMaximumBinaryTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	m, im := nums[0], 0
	for i, n := range nums {
		if n > m {
			m, im = n, i
		}
	}
	return &TreeNode{
		Val:   m,
		Left:  constructMaximumBinaryTree(nums[:im]),
		Right: constructMaximumBinaryTree(nums[im+1:]),
	}
}