标签：diff arch present res balance == fine int efi
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 == null || nums.length == 0) { return null; } return buildBST(0, nums.length - 1, nums); } private TreeNode buildBST(int start, int end, int[] nums) { if (start > end) { return null; } int mid = start + (end - start) / 2; TreeNode cur = new TreeNode(nums[mid]); cur.left = buildBST(start, mid - 1, nums); cur.right = buildBST(mid + 1, end, nums); return cur; } }
[LC] 108. Convert Sorted Array to Binary Search Tree
标签：diff arch present res balance == fine int efi
原文地址：https://www.cnblogs.com/xuanlu/p/12170624.html
