算法leetcode｜面试题 04.02. 最小高度树（rust重拳出击）

Posted 2022-11-30 二当家的白帽子

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文章目录

• 面试题 04.02. 最小高度树：
• • 样例 1：
  • 原题传送门：
• 分析
• 题解
• • rust
  • go
  • c++
  • java
  • typescript
  • python

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面试题 04.02. 最小高度树：

给定一个有序整数数组，元素各不相同且按升序排列，编写一个算法，创建一棵高度最小的二叉搜索树。

样例 1：

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 

原题传送门：

https://leetcode.cn/problems/minimum-height-tree-lcci/

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分析

• 面对这道算法题目，二当家的陷入了沉思。
• 最小高度，就意味着要尽可能平衡，完全，满。
• 二叉搜索树本来是有一些限制的，但是给的数组已经排好序了，那我们每次从中间分成左右子树，简直完美。

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题解

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 
    pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> 
        fn dfs(nums: &Vec<i32>, l: i32, r: i32) -> Option<Rc<RefCell<TreeNode>>> 
            if l > r 
                return Option::None;
            

            let mid = (l + r) >> 1;
            // let mid = (l + r + 1) >> 1;
            let mut root = TreeNode::new(nums[mid as usize]);
            root.left = dfs(nums, l, mid - 1);
            root.right = dfs(nums, mid + 1, r);

            return Option::Some(Rc::new(RefCell::new(root)));
        

        return dfs(&nums, 0, (nums.len() - 1) as i32);
    

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go

/**
 * Definition for a binary tree node.
 * type TreeNode struct 
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * 
 */
func sortedArrayToBST(nums []int) *TreeNode 
    var dfs func([]int, int, int) *TreeNode
	dfs = func(nums []int, l int, r int) *TreeNode 
		if l > r 
			return nil
		

		mid := (l + r) >> 1
		//mid := (l + r + 1) >> 1
		return &TreeNode
			nums[mid],
			dfs(nums, l, mid-1),
			dfs(nums, mid+1, r),
		
	

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

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c++

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) 
 * ;
 */
class Solution 
private:
    TreeNode* dfs(vector<int>& nums, int l, int r) 
        if (l > r) 
            return nullptr;
        

        int mid = (l + r) >> 1;
//        int mid = (l + r + 1) >> 1;
        TreeNode *root = new TreeNode(nums[mid]);
        root->left = dfs(nums, l, mid - 1);
        root->right = dfs(nums, mid + 1, r);
        return root;
    
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) 
        return dfs(nums, 0, nums.size() - 1);
    
;

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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) 
        return dfs(nums, 0, nums.length - 1);
    

    private TreeNode dfs(int[] nums, int l, int r) 
        if (l > r) 
            return null;
        

        int mid = (l + r) >> 1;
        // int mid = (l + r + 1) >> 1;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = dfs(nums, l, mid - 1);
        root.right = dfs(nums, mid + 1, r);
        return root;
    

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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 dfs = function (nums: number[], l: number, r: number): TreeNode | null 
		if (l > r) 
			return null;
		

		const mid = (l + r) >> 1;
		//const mid = (l + r + 1) >> 1;
		return new TreeNode(
			nums[mid],
			dfs(nums, l, mid - 1),
			dfs(nums, mid + 1, r)
		);
	;

	return dfs(nums, 0, nums.length - 1);
;

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python

# 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:
        def dfs(l: int, r: int) -> TreeNode:
            if l > r:
                return None
            mid = (l + r) >> 1
            # mid = (l + r + 1) >> 1
            return TreeNode(
                nums[mid],
                dfs(l, mid - 1),
                dfs(mid + 1, r),
            )

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

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