Convert Sorted Array to Binary Search Tree

将排序数组转换为平衡二叉搜索树

最新推荐文章于 2022-02-19 22:56:38 发布

weixin_30872157

最新推荐文章于 2022-02-19 22:56:38 发布

阅读量69

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zengzy/p/5056210.html

本文介绍了一种方法，通过递归将已排序的整数数组转换为高度平衡的二叉搜索树（BST）。这种方法确保了每个节点的左右子树的高度之差不超过1，从而提高了树的效率。

Convert Sorted Array to Binary Search Tree

Total Accepted: 61516 Total Submissions: 173462 Difficulty: Medium

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

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