[LeetCode] Convert Sorted Array to Binary Search Tree

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

[Analysis]
注意：题目中的数组是已经排好序的，如果没有排序，则需要用BST的左旋、右旋、左右旋、右左旋等进行调整。

对于已经排好序的数组，构造BST时，使用数组的中间元素array[mid]作为BST的根节点，array[0:mid-1]用于构造其左子树，array[mid+1:end]用于构造其右子树，这两步可以递归进行。

[Solution]

// Definition for binary tree
// struct TreeNode {
//     int val;
//     TreeNode *left;
//     TreeNode *right;
//     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
// };

class Solution {
public:
    // build tree
    TreeNode *build(vector &num, int begin, int end){
        if(begin > end){
            return NULL;
        }
   
        // build left brunch and right brunch
        int mid = (begin+end)/2;
        TreeNode* left = build(num, begin, mid-1);
        TreeNode* right = build(num, mid+1, end);
   
        // build tree
        TreeNode* node = new TreeNode(num[mid]);
        node->left = left;
        node->right = right;
        return node;
    }
   
    // sorted array to BST
    TreeNode *sortedArrayToBST(vector &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(num.size() == 0){
            return NULL;
        }
        else{
            return build(num, 0, num.size()-1);
        }
    }
};

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