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

1. 中间的index不是除以2，而是mid = low + (high - low) / 2;

2. 初始化节点： TreeNode node = new TreeNode(0);
   和改变节点：node = helper(nums, 0, nums.length - 1);
   是两个步骤的时候，就要用helper函数

3. 好吧，原来高度不超过1是通过index来控制的，也是把左右当做参数：左边一个右边一个

 

node.left =
node.right =
都是在helper中完成的，helper可以多功能一些
helper中也有边界条件

而且用了数组的话，函数的参数最好带上数组名

 

自己写的，不知道里面为啥会链表有环

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        //特殊情况
        if (nums.length < 0 || nums == null) 
            return null;
        
        TreeNode root = new TreeNode(0);
        return helper(root, nums, 0, nums.length - 1);
    }
    
    TreeNode helper(TreeNode node, int[] nums, int left, int right) {
        //特殊情况
        if (left > right) {
            return null;
        }
        
        int mid = left + (right - left) / 2;
        node.left = helper(node, nums, left, mid - 1);
        node.right = helper(node, nums, mid + 1, right);
            
        return node;
    }
}

 正确版本

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        //特殊情况
        if (nums.length < 0 || nums == null) 
            return null;
        
        TreeNode root = new TreeNode(0);
        return helper(nums, 0, nums.length - 1);
    }
    
    TreeNode helper(int[] nums, int left, int right) {
        //特殊情况
        if (left > right) {
            return null;
        }
        
        int mid = left + (right - left) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(nums, left, mid - 1);
        node.right = helper(nums, mid + 1, right);
            
        return node;
    }
}