108 Convert Sorted Array to Binary Search Tree数组变成高度平衡的二叉树

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

复习了还是不会的地方:
中间的怎么选取啊?int mid = low + (high - low) / 2;
然后可以加个helper函数

/**
 * 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) {
        TreeNode node = new TreeNode(0);
        //corner case
        if (nums.length == 0) {
            return node;
        }
        node = helper(nums, 0, nums.length - 1);// -1 should be noticed ahead
        return node;
    }
    
    public TreeNode helper(int[] nums, int low, int high) {      
    //corner case : low > high
        if (low > high) {
            return null;
        }
        int mid = low + (high - low) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        
        root.left = helper(nums, low, mid - 1);
        root.right = helper(nums, mid + 1, high);
        
        return root;
    }
}
 
原文地址:https://www.cnblogs.com/immiao0319/p/14984301.html