108. 将有序数组转换为二叉搜索树 (opens new window)

English Version (opens new window)

题目描述

给你一个整数数组 nums ，其中元素已经按 升序 排列，请你将其转换为一棵 高度平衡 二叉搜索树。

高度平衡 二叉树是一棵满足「每个节点的左右两个子树的高度差的绝对值不超过 1 」的二叉树。

示例 1：

输入：nums = [-10,-3,0,5,9]
输出：[0,-3,9,-10,null,5]
解释：[0,-10,5,null,-3,null,9] 也将被视为正确答案：

示例 2：

输入：nums = [1,3]
输出：[3,1]
解释：[1,3] 和 [3,1] 都是高度平衡二叉搜索树。

提示：

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums 按 严格递增 顺序排列

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        def buildBST(nums, start, end):
            if start > end:
                return None
            mid = (start + end) >> 1
            return TreeNode(nums[mid], buildBST(nums, start, mid - 1), buildBST(nums, mid + 1, end))

        return buildBST(nums, 0, len(nums) - 1)

Java

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return buildBST(nums, 0, nums.length - 1);
    }

    private TreeNode buildBST(int[] nums, int start, int end) {
        if (start > end) {
            return null;
        }
        int mid = (start + end) >> 1;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = buildBST(nums, start, mid - 1);
        root.right = buildBST(nums, mid + 1, end);
        return root;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return buildBST(nums, 0, nums.size() - 1);
    }

private:
    TreeNode* buildBST(vector<int>& nums, int start, int end) {
        if (start > end) {
            return nullptr;
        }
        int mid = (start + end) / 2;
        TreeNode *root = new TreeNode(nums[mid]);
        root->left = buildBST(nums, start, mid - 1);
        root->right = buildBST(nums, mid + 1, end);
        return root;
    }
};

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function(nums) {
    const buildBST = (nums, start, end) => {
        if (start > end) {
            return null;
        }
        const mid = (start + end) >> 1;
        const root = new TreeNode(nums[mid]);
        root.left = buildBST(nums, start, mid - 1);
        root.right = buildBST(nums, mid + 1, end);
        return root;
    }

    return buildBST(nums, 0, nums.length - 1);
};

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