LeetCode 108: Convert Sorted Array to Binary Search Tree

Convert Sorted Array to Binary Search Tree

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

解题思路
为了得到高度平衡的二叉树，选中点 $middle = (start + end)/2$ 构造根节点，然后递归的构造左子树和右子树。代码如下：

/**
 * 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 *sortedArrayToBST(vector<int>& nums, int start, int end) {
        if (start >= end) return NULL;

        int middle = (start + end) >> 1; // 中间元素

        TreeNode *root = new TreeNode(nums[middle]);
        root->left = sortedArrayToBST(nums, start, middle);
        root->right = sortedArrayToBST(nums, middle + 1, end);

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