[leetcode]108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2025-08-15 16:44:13 发布

你看见我的代码了么

最新推荐文章于 2025-08-15 16:44:13 发布

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本文介绍如何将一个升序排列的数组转换为高度平衡的二叉搜索树。平衡二叉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。通过递归方式选择中间元素作为根节点，实现树的平衡。

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[leetcode]108. Convert Sorted Array to Binary Search Tree

Analysis

组会又被怼了—— [哎~]

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.
这题是用递归法构建一个平衡二叉树
平衡二叉树：a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Implement

/**
 * 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 {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if(nums.size()==0)
            return NULL;
        return myBST(nums, 0, nums.size()-1);
    }
    TreeNode* myBST(vector<int>& nums, int l, int r){
        if(l > r)
            return NULL;
        int mid = (l+r)/2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = myBST(nums, l, mid-1);
        root->right = myBST(nums, mid+1, r);
        return root;
    }
};