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

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要点就是找到中心点,然后分别递归构造左边的数和右边的数

TreeNode* sortedArrayToBST(vector<int>& nums, int beg, int end) {
if (beg > end)
return nullptr;
int mid = (beg + end) >> ;
TreeNode* root = new TreeNode(nums[mid]);
root->left = sortedArrayToBST(nums, beg, mid - );
root->right = sortedArrayToBST(nums, mid + , end);
return root; } TreeNode* sortedArrayToBST(vector<int>& nums) {
return sortedArrayToBST(nums, , nums.size()-);
}

链表为了代码的简洁,我使用了二级指针,不过可读性变差了

TreeNode* sortedListToBST(ListNode* head) {
if (head == nullptr)
return nullptr;
ListNode* fast = head;
ListNode** slow = &head;
while (fast->next != nullptr && fast->next->next != nullptr)
{
fast = fast->next->next;
slow = &((*slow)->next);
}
TreeNode* root = new TreeNode((*slow)->val); root->right = sortedListToBST((*slow)->next);
*slow = nullptr;
root->left = sortedListToBST(head); return root;
}

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