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

Analysis

因为数组是排序过的,因此只需要利用递归,将中点设为根节点,左右子树则分别对应被中点分隔的左右两部分有序数组,再递归生成子树即可。

Solution

/**
* 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(nums, 0, nums.length - 1);
}
TreeNode buildBST(int[] nums, int left, int right) {
if (left > right) {
return null;
}
int mid = (left + right) / 2;
TreeNode node = new TreeNode(nums[mid]);
node.left = buildBST(nums, left, mid - 1);
node.right = buildBST(nums, mid + 1, right);
return node;
}
}