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.

最大二叉树。

给定一个不含重复元素的整数数组。一个以此数组构建的最大二叉树定义如下：

二叉树的根是数组中的最大元素。
左子树是通过数组中最大值左边部分构造出的最大二叉树。
右子树是通过数组中最大值右边部分构造出的最大二叉树。
通过给定的数组构建最大二叉树，并且输出这个树的根节点。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题的题目描述，就是在暗示其实是可以用递归的思路做的。我们用一个函数找到数组中的最大值及其index，然后用分治的思路递归去构建左子树和右子树。

时间O(nlogn) - average; O(n^2) - worst case

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode constructMaximumBinaryTree(int[] nums) {
18         return helper(nums, 0, nums.length);
19     }
20 
21     private TreeNode helper(int[] nums, int left, int right) {
22         if (left == right) {
23             return null;
24         }
25         int maxIndex = max(nums, left, right);
26         TreeNode root = new TreeNode(nums[maxIndex]);
27         root.left = helper(nums, left, maxIndex);
28         root.right = helper(nums, maxIndex + 1, right);
29         return root;
30     }
31 
32     private int max(int[] nums, int left, int right) {
33         int maxIndex = left;
34         for (int i = left; i < right; i++) {
35             if (nums[maxIndex] < nums[i]) {
36                 maxIndex = i;
37             }
38         }
39         return maxIndex;
40     }
41 }

 

这道题还有一个类似单调栈的做法但是很不好想，我把引用贴在这里，有兴趣可以看一下，运行速度其实并没有非常快。

LeetCode 题目总结