g0601_0700.s0654_maximum_binary_tree.readme.md Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-kotlin Show documentation
Show all versions of leetcode-in-kotlin Show documentation
Kotlin-based LeetCode algorithm problem solutions, regularly updated
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**.