5.3.4 Convert Sorted Array to Binary Searh Tree

ZhikangFu

于 2015-06-09 15:49:29 发布

阅读量410

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/ZhikangFu/article/details/46427635

leetcode 专栏收录该内容

54 篇文章

订阅专栏

本文介绍了一种将有序数组转换为高度平衡的二叉搜索树的方法。通过递归构建二叉树，每次选择中间元素作为根节点，确保左右子树尽可能平衡。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Notes:
 Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
 Solution: Recursion.
 */

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        return buildBST(num, 0, num.size() - 1);
    }
    
    TreeNode *buildBST(vector<int> &num, int start, int end)
    {
        if (start > end) return NULL;

        int mid = (start + end) / 2;
        TreeNode *root = new TreeNode(num[mid]);
        root->left = buildBST(num, start, mid - 1);
        root->right = buildBST(num, mid + 1, end);

        return root;
    }
};