Easy
Given an integer array nums
where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.
A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.
Example 1:
Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:
Example 2:
Input: nums = [1,3]
Output: [3,1]
Explanation: [1,3] and [3,1] are both a height-balanced BSTs.
Constraints:
1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums
is sorted in a strictly increasing order.import com_github_leetcode.TreeNode;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public TreeNode sortedArrayToBST(int[] num) {
/*1. Set up recursion*/
return makeTree(num, 0, num.length - 1);
}
private TreeNode makeTree(int[] num, int left, int right) {
/*2. left as lowest# can't be greater than right*/
if (left > right) {
return null;
}
/*3. Set median# as node*/
int mid = (left + right) / 2;
TreeNode midNode = new TreeNode(num[mid]);
/*4. Set mid node's kids*/
midNode.left = makeTree(num, left, mid - 1);
midNode.right = makeTree(num, mid + 1, right);
/*5. Sends node back || Goes back to prev node*/
return midNode;
}
}