[LeetCode] 108. Convert Sorted Array to Binary Search Tree

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

本题要求把有序数组转化成二叉搜索树。二叉搜索树满足性质 左<根<右，对二叉搜索树进行中序遍历将得到有序数组，
因此有序数组的中间值为根结点，前半部分为左子树，后半部分为右子树，各子树均符合该特性。所以使用二分法的方法进行求解。

代码如下：

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