[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.

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

Solution:

/**
* 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) {
// 边界
if (nums.size() == 0) return NULL;
if (nums.size() == 1) return new TreeNode(nums[0]);

// 每次取子数组的中位数作为根节点值
int mid = nums.size() / 2;
int rootVal = nums[mid];
TreeNode* root = new TreeNode(rootVal);

// 拆分左右子数组
vector<int> leftSubArray(nums.begin(), nums.begin() + mid);
vector<int> rightSubArray(nums.begin() + mid + 1, nums.end());

root->left = sortedArrayToBST(leftSubArray); //dfs构造左子树
root->right = sortedArrayToBST(rightSubArray); //dfs构造右子树

return root;
}
};
解法参考自：https://discuss.leetcode.com/topic/6472/accepted-c-recursive-solution-within-a-single-method