给你一个整数数组 nums ，其中元素已经按 升序排列，请你将其转换为一棵 高度平衡二叉搜索树。

高度平衡二叉树是一棵满足「每个节点的左右两个子树的高度差的绝对值不超过 1 」的二叉树。

提示：

• 1 <= nums.length <= $10^4$
• $-10^4$ <= nums[i] <= $10^4$
• nums 按严格递增顺序排列

题解1 DFS构建二叉搜索树（平衡树）【前序】

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* build(vector<int>& nums, int left, int right){
        if(left > right) return nullptr;
        
        int mid = (left + right) >> 1;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = build(nums, left, mid-1);
        root->right = build(nums, mid+1, right);
        return root;

    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        // 含right
        return build(nums, 0, nums.size()-1);
    }
};

题解2 中序

class Solution {
    int index = 0;
public:
    TreeNode* build(vector<int>& nums, int left, int right){
        if(left > right) return nullptr;

        int mid = (left + right) >> 1;
        // 中序（操作放中间）
        TreeNode* root = new TreeNode();
        root->left = build(nums, left, mid-1);
        root->val = nums[index++];
        root->right = build(nums, mid+1, right);
        return root;

    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        // 含right
        return build(nums, 0, nums.size()-1);
    }
};