we could use Divide and Conquer.
we could do it recursively
pick the middle element as the root
then pass the left to the sortedArrayToBST, the array is from 0 to mid-1
pass the right to sortedArrayToBST, the array is from mid+1 the end
when the str is empty, return NULL
Several minor mistakes:
1. When debug in eclipse, you should always check if you have include the corresponding headers, or you would got strange errors(error: expected ‘;’ at end of member declaration).
2. DO NOT FORGET THE RETURN VALUE IN IF-ELSE CONDITION!
96 milli secs to pass Judge Large
class Solution
{
public:
TreeNode* sortedArrayToBSTHelper(int* arr, int start, int end)
{
if (start <= end)
{
int len = end - start;
int mid = start + len / 2;
TreeNode* root = new TreeNode(arr[mid]);
root->left = sortedArrayToBSTHelper(arr, start, mid - 1);
root->right = sortedArrayToBSTHelper(arr, mid + 1, end);
return root;
}
else
return NULL;
};
TreeNode* sortedArrayToBST(vector<int> &num)
{
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (num.empty())
return NULL;
int arr[num.size()];
for (int i = 0; i < num.size(); i++)
{
arr[i] = num[i];
}
return sortedArrayToBSTHelper(arr, 0, num.size() - 1);
};
};