Given a singly linked list 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 linked list: [-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
思路:
两种方案,一种偷懒的方案,先转换为数组再转二叉树,另一种是利用生成树的先后顺序进行二叉树的构造。
1 偷懒版
TreeNode* sortedArrayToBST(vector<int>& nums) {
if (nums.size() == 0) return NULL;
int mid = nums.size() / 2;
TreeNode *root = new TreeNode(nums[mid]);
vector<int> leftNums(nums.begin(), nums.begin() + mid);
vector<int> rightNums(nums.begin() + mid + 1, nums.end());
root->left = sortedArrayToBST(leftNums);
root->right = sortedArrayToBST(rightNums);
return root;
}
TreeNode* sortedListToBST(ListNode* head) {
vector<int> nums;
ListNode *curr = head;
while (curr){
nums.push_back(curr->val);
curr = curr->next;
}
return sortedArrayToBST(nums);
}
2 改变顺序版
TreeNode *sortedListToBST2(int left, int right, ListNode *&head){
if (!head || left > right) return NULL;
int mid = left + (right - left) / 2;
TreeNode *leftNode = sortedListToBST2(left, mid - 1, head);
TreeNode *root = new TreeNode(head->val);
head = head->next;
TreeNode *rightNode = sortedListToBST2(mid + 1, right, head);
root->left = leftNode;
root->right = rightNode;
return root;
}
TreeNode* sortedListToBST2(ListNode* head) {
int len = 0;
ListNode *curr = head;
while (curr){
len++;
curr = curr->next;
}
return sortedListToBST2(0, len, head);
}