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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums.length == 0) return null;
        return help(nums,0, nums.length - 1);
    }
    public TreeNode help(int[] nums, int start, int end){
        if(end < start) return null;
        int rootind = (end + start) / 2;
        TreeNode root = new TreeNode(nums[rootind]);
        root.left = help(nums, start, rootind - 1);
        root.right = help(nums, rootind + 1, end);
        return root;
    }
}

没什么特殊的，每一步都是把中间的当作root，然后两边作为subtree。

递归终止条件是end<start，因为存在==的情况（单一node）