leetcode 108.Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

给一个升序数组，让重建一个AVL二叉搜索树

思路：
AVL就是左右子树高度差不能>1
$N_h$表示高度为h的node数

$N_{h-1},N_{h-2}$表示左子树和右子树（不按顺序）

$N_h=1+N_{h-1}+N_{h-2}>2N_{h-2}$
其中$N_h$表示高度为h的node数

为了保持平衡，尽量左子树右子树node数一样

而已经知道BST中序遍历得到的是升序数组
中序遍历顺序：左子树，root，右子树

所以取中间元素作root
然后递归

//0ms
public TreeNode sortedArrayToBST(int[] nums) {
    
    return buildTree(nums, 0, nums.length-1);
}

TreeNode buildTree(int[] nums, int start, int end) {
    if(start > end) return null;
    
    int rootIdx = start + (end - start) / 2;
    TreeNode root = new TreeNode(nums[rootIdx]);
    root.left = buildTree(nums, start, rootIdx-1);
    root.right = buildTree(nums, rootIdx+1, end);
    
    return root;
}