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108. 将有序数组转换为二叉搜索树

English Version

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,null,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 >> 1;
        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);
};

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function sortedArrayToBST(nums: number[]): TreeNode | null {
    const n = nums.length;
    if (n === 0) {
        return null;
    }
    const mid = n >> 1;
    return new TreeNode(
        nums[mid],
        sortedArrayToBST(nums.slice(0, mid)),
        sortedArrayToBST(nums.slice(mid + 1)),
    );
}

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
	return buildBST(nums, 0, len(nums)-1)
}

func buildBST(nums []int, start, end int) *TreeNode {
	if start > end {
		return nil
	}
	mid := (start + end) >> 1
	return &TreeNode{
		Val:   nums[mid],
		Left:  buildBST(nums, start, mid-1),
		Right: buildBST(nums, mid+1, end),
	}
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn to_bst(nums: &Vec<i32>, start: usize, end: usize) -> Option<Rc<RefCell<TreeNode>>> {
        if start >= end {
            return None;
        }
        let mid = start + (end - start) / 2;
        Some(Rc::new(RefCell::new(TreeNode {
            val: nums[mid],
            left: Self::to_bst(nums, start, mid),
            right: Self::to_bst(nums, mid + 1, end),
        })))
    }

    pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::to_bst(&nums, 0, nums.len())
    }
}

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