【Lintcode】106.Convert Sorted List to Balanced BST-优快云博客

本文介绍如何将一个按升序排列的单链表转换为高度平衡的二叉搜索树（BST）。提供了两种解决方案，包括使用快慢指针进行中点查找的方法，以及递归地将链表划分并构建树节点。

题目：

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

Example

               2
1->2->3  =>   / \
             1   3

题解：

Solution 1 ()

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if (!head) return nullptr;
        
        return sortedListToBST(head, nullptr); 
    }
    TreeNode *sortedListToBST(ListNode* head, ListNode* tail) {
        if (head == tail) return nullptr;
        ListNode* mid = head, *tmp = head;
        
        while (tmp != tail && tmp->next != tail) {
            mid = mid->next;
            tmp = tmp->next->next;
        }
        TreeNode* root = new TreeNode(mid->val);
        root->left = sortedListToBST(head, mid);
        root->right = sortedListToBST(mid->next, tail);
        
        return root;
    }

};

Solution 2 ()

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

};