[LeetCode] 108. 将有序数组转换为二叉搜索树

题目链接 : https://leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree/

题目描述:

将一个按照升序排列的有序数组，转换为一棵高度平衡二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

给定有序数组: [-10,-3,0,5,9],

一个可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面这个高度平衡二叉搜索树：

      0
      / \
    -3   9
   /   /
 -10  5

思路:

我们找到数组的中点, 然后分成两部分,

比如 [-10,-3,0,5,9], 节点0的左边[-10, -3],右边[5, 9]

依次递归下去.

代码:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        if not nums: return
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid+1:])
        return root

java

/**
 * 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.length == 0) return null;
        return helper(nums, 0, nums.length - 1);
    }

    private TreeNode helper(int[] nums, int left, int high) {
        if (left > high) return null;
        int mid = left + (high - left) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(nums, left, mid - 1);
        root.right = helper(nums, mid + 1, high);
        return root;
    }
}