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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
Analyse: Using binary search to divide the sorted array into two parts, then recursively do the same process for the two parts.
Runtime: 16ms.
Additional: IT‘S THE FIRST TIME I COME UP WITH A SOLUTION WITHOUT REFERENCE. CONG!!
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode* sortedArrayToBST(vector<int>& nums) { 13 if(nums.size() == 0) return NULL; 14 15 return convert(nums, 0, nums.size() - 1); 16 } 17 TreeNode* convert(vector<int>& nums, int low, int high){ 18 if(low > high) return NULL; 19 20 int mid = (low + high) / 2; 21 TreeNode* root = new TreeNode(nums[mid]); 22 if(low == high) return root; 23 24 root->left = convert(nums, low, mid - 1); 25 root->right = convert(nums, mid + 1, high); 26 27 return root; 28 } 29 30 };
Convert Sorted Array to Binary Search Tree
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原文地址：http://www.cnblogs.com/amazingzoe/p/4691506.html
