#Leetcode# 108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/10110675.html

本文介绍了一种将已排序的数组转换为高度平衡的二叉搜索树的方法。平衡二叉搜索树是一种特殊的二叉树，其中每个节点的两个子树的深度相差不超过1。通过递归地选取数组中间元素作为根节点，可以确保生成的树尽可能平衡。

https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

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

代码：

/**
 * 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 {
	TreeNode * BuildTree(vector<int> &nums, int start, int end)
	{
		if(start >= end)
			return NULL;
		int mid = (start + end)/2;
		TreeNode * cur = new TreeNode(nums[mid]);
		cur ->left = BuildTree(nums, start, mid);
		cur ->right = BuildTree(nums, mid + 1, end);

		return cur;
	}
public:
	TreeNode* sortedArrayToBST(vector<int>& nums) {
        int n = nums.size();
		return BuildTree(nums, 0, n);
	}
};

　　不是很明白为什么是这么写的本来想的是写一个 $insert$ 函数然后跑 $nums$ 数组但是没写出来肥去之后再想吧指针指来指去不是很搞得清

转载于:https://www.cnblogs.com/zlrrrr/p/10110675.html