LeetCode Brush title notes - recursion - to convert to an ordered array of binary search trees

Title Description

An ordered array in accordance with the ascending order, is converted to a highly balanced binary search tree.

In this problem, a highly balanced binary tree is a binary tree refers to the left and right sub-tree of each node is the height difference between the absolute value of not more than 1.

Example:

Given an ordered array: [-10, -3,0,5,9],

One possible answer is: [0, -3,9, -10, null, 5], it can be expressed below this height balanced binary search tree:

0
/ \
-3 9
/ /
-10 5

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree

Problem-solving ideas:

1, the root root = nums [mid]

2, root.left = intermediate value of the remaining portion of the left

3, root.right = intermediate value of the remaining portion of the left

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode sortedArrayToBST(int[] nums) {
12         if(nums.length==0) return null;
13         else{
14             return digui(nums,0,nums.length-1);
15         }
16     }
17     
18     public TreeNode digui(int[] nums,int left,int right){
19         if(left>right){
20             return null;
21         }
22         
23         int mid=left+(right-left)/2;  //避免溢出风险；；；；
24         TreeNode root = new TreeNode(nums[mid]);
25         root.left=digui(nums,left,mid-1);
26         root.right=digui(nums,mid+1,right);
27         return root;
28     }
29 }

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