0654. Maximum Binary Tree
https://leetcode.com/problems/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:
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.
ac
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
return build(nums, 0, nums.length-1);
}
public TreeNode build(int[] nums, int start, int end) {
if (start > end) return null;
// find max
int maxIdx = -1, max = Integer.MIN_VALUE;
for (int i = start; i <= end; i++) {
if (nums[i] > max) {
max = nums[i];
maxIdx = i;
}
}
// build
TreeNode root = new TreeNode(max);
root.left = build(nums, start, maxIdx-1);
root.right = build(nums, maxIdx+1, end);
return root;
}
}
/*
just recursion build the subtree
*/
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