【LeetCode】Convert Sorted List to Binary Search Tree

题目描述：

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

将一个单向链表转换成二叉搜索树。

我的解题思路是用AVL树插入的方法来将链表中的元素一个个接上。因为链表是已经排序好的，每一个节点都大于前一个节点，因此新插入的节点都位于最右下的位置，同样进行旋转操作时只需要进行一次左旋就可以了。用栈来记录下到达最右下节点的路径，方便寻找父节点；用map来记录下路径上每个节点对应的左孩子深度，右孩子深度新插入的节点为0，每向上爬一个父节点则++。要注意爬到根时单独处理。

代码如下：

//Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};//Definition for binary treestruct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:TreeNode *sortedListToBST(ListNode *head) {if (!head)return NULL;TreeNode *tNode = new TreeNode(head->val);TreeNode *root = tNode;stack<TreeNode*> path;//路径unordered_map<TreeNode*, int> ldepth;//与path对应节点的左孩子深度path.push(tNode);ldepth[tNode] = 0;head = head->next;while (head){TreeNode *front = path.top();//寻找最右下节点while (front->right){path.push(front->right);front = front->right;}//将新值接到右下节点,该节点左孩子深度为0front->right = new TreeNode(head->val);ldepth[front->right] = 0;//向上寻找最小非平衡二叉树int currright = 1;TreeNode *next(NULL);while (abs(ldepth[front] - currright) <= 1 && !path.empty()){next = front;path.pop();if (!path.empty())front = path.top();currright++;}//如果path为空，即该树仍然是平衡二叉树，则continue//否则向左旋转if (path.empty()){head = head->next;path.push(root);continue;}TreeNode *curr = front;path.pop();if (path.empty())root = next;else{front = path.top();front->right = next;}curr->right = next->left;next->left = curr;ldepth[next] = ldepth[curr] + 1;path.push(next);head = head->next;}return root;}};