LeetCode 108. Convert Sorted Array to Binary Search Tree

Solution1：我的答案
构造二叉树利用递归

/*** 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<int>& nums) {return my_sort(nums);}TreeNode* my_sort(vector<int> nums) {if (nums.size() == 0) return NULL;else if (nums.size() == 1) {TreeNode* root = new TreeNode(nums[0]);return root;}else {int root_index = nums.size()/2;TreeNode* root = new TreeNode(nums[root_index]);vector<int> left_tree, right_tree;for (int i = 0; i < nums.size(); i++) {if (i < root_index)left_tree.push_back(nums[i]);else if (i > root_index)right_tree.push_back(nums[i]);}root->left = my_sort(left_tree);root->right = my_sort(right_tree);return root;}}
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4295245.html
下面是避免数组元素大量复制的写法，还是这种好啊

class Solution {
public:TreeNode *sortedArrayToBST(vector<int> &num) {return sortedArrayToBST(num, 0 , num.size() - 1);}TreeNode *sortedArrayToBST(vector<int> &num, int left, int right) {if (left > right) return NULL;int mid = (left + right) / 2;TreeNode *cur = new TreeNode(num[mid]);cur->left = sortedArrayToBST(num, left, mid - 1);cur->right = sortedArrayToBST(num, mid + 1, right);return cur;}
};

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