［LeetCode］Convert Sorted Array to Binary Search Tree

解题思路：
1，对一个有序的数组，其中间那个数 是 整颗树的根；
2，利用递归的思想，重复1的操作；

public class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return sortedArrayToBST(nums, 0, nums.length);
}

private TreeNode sortedArrayToBST(int[] nums, int start, int end){
if (start >= end) return null;
int mid = (start + end) / 2;
TreeNode node = new TreeNode(nums[mid]);
node.left = sortedArrayToBST(nums, start, mid);
node.right = sortedArrayToBST(nums, mid+1, end);

return node;
}
}

-------------------------------第一遍解题思路－－－－－－－－－－－－－－－－
解题思路：
1，计算height
2，计算leafnumber
3，利用所有最底层的leaf都as left as posiible的特点

/**
* 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) {

TreeNode* root = NULL;
int height = log2(nums.size());
int count = 0;
for(int i = 0; i < height; ++i){
count += pow(2, i);
}
int leafNumber = nums.size() - count;

int index = 0;
build(root, height, leafNumber, nums, index);

return root;

}

void build(TreeNode* &root, int height, int &leafNumber, vector<int>& nums, int &index){
if(index == nums.size()) return ;
if (height == 0){
if (leafNumber > 0){
leafNumber--;
root = new TreeNode(nums[index]);
index++;
}
}else{
root = new TreeNode(0);
build(root->left, height-1, leafNumber, nums, index);
root->val = nums[index++];
build(root->right, height-1, leafNumber, nums, index);
}
return;
}
};