convert-sorted-array-to-binary-search-tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
解法一：

/** * 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) {        int low=0,high=num.size()-1;        TreeNode *root=dfs(low,high,num);        return root;    }    TreeNode* dfs(int low,int high,vector<int> &num)    {        if(low>high)            return NULL;        else        {            int mid=low+(high-low+1)/2;//奇数时，取最中间的值；偶数时，取偏后的那个中间值（因为只有这样才能ac，这是测试用例设计的问题；若是单存的抛开测试用例的话，取靠前的那个中间值也是对的）            TreeNode *root=new TreeNode(num[mid]);            root->left=dfs(low,mid-1,num);            root->right=dfs(mid+1,high,num);            return root;        }            }};

解法二：

/** * 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) {        int low=0,high=num.size()-1;        TreeNode *root=NULL;        dfs(root,low,high,num);        return root;    }   void dfs(TreeNode *&root,int low,int high,vector<int> &num)    {        if(low>high)            return;        else        {            int mid=low+(high-low+1)/2;//奇数时，取最中间的值；偶数时，取偏后的那个中间值（因为只有这样才能ac，这是测试用例设计的问题）            root=new TreeNode(num[mid]);            dfs(root->left,low,mid-1,num);            dfs(root->right,mid+1,high,num);        }    }};