Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
解题思路:
1. 找到数组的中间节点将其置为根节点
2. 左边的即为左子树
3. 右边的即为右子树
4. 递归求解
/**
* 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* sortBST(vector<int>& nums, int begin, int end){
if(begin > end)
return 0;
int rootIndex = begin+(end-begin)/2; //更简单的应该是 (begin+end)/2
TreeNode* root = new TreeNode(nums[rootIndex]);
root->left = sortBST(nums, begin, rootIndex-1);
root->right = sortBST(nums, rootIndex+1, end);
return root;
}
TreeNode* sortedArrayToBST(vector<int>& nums) {
return sortBST(nums, 0, nums.size()-1);
}
};