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
二叉搜索樹(BST)
- 若任意節(jié)點(diǎn)的左子樹不空讲衫，則左子樹上所有結(jié)點(diǎn)的值均小于它的根結(jié)點(diǎn)的值；
- 若任意節(jié)點(diǎn)的右子樹不空孵班，則右子樹上所有結(jié)點(diǎn)的值均大于它的根結(jié)點(diǎn)的值涉兽；
- 它是一棵空樹或它的左右兩個(gè)子樹的高度差的絕對(duì)值不超過1，并且左右兩個(gè)子樹都是一棵平衡二叉樹
/**
* 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) {
TreeNode root = treeHelper(nums, 0, nums.length-1);
return root;
}
public TreeNode treeHelper(int[] nums, int left, int right){
if (left > right){
return null;
}
int mid = (left + right)/2;
TreeNode root = new TreeNode(nums[mid]);
root.left = treeHelper(nums, left, mid - 1);
root.right = treeHelper(nums, mid + 1, right);
return root;
}
}
