leetcode - Convert Sorted Array to Binary Search Tree

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

将一个升序数组建立一个高度平衡的二叉查找树。 如果高度平衡，则根节点一定是中间的数。用递归。