leetcode 109. Convert Sorted List to Binary Search Tree 链表构造平衡二叉搜索树 + DFS

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

根据一个排序好的链表构造二叉平衡树，这个问题的重点是找到中间元素，使用双指针即可。

建议和 leetcode 108. Convert Sorted Array to Binary Search Tree 构建平衡二叉搜索树 + DFS、leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST、 leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal 中前序构造BST放到一起学习。

代码如下：

/*class ListNode {     int val;     ListNode next;     ListNode(int x) { val = x; }}class TreeNode{     int val;     TreeNode left;     TreeNode right;     TreeNode(int x) { val = x; }}*//* * 这个题很简单，主要就是使用数组类似的方法去做 * 问题的关键就是找到middle位置，可以考虑使用双指针来解决 * */public class Solution {    public TreeNode sortedListToBST(ListNode head)     {        return toBST(head,null);    }    public TreeNode toBST(ListNode head, ListNode tail)    {        if(head == tail)            return null;        else        {            ListNode mid = getMid(head,tail);            TreeNode root = new TreeNode(mid.val);            //注意这个要考虑到head和tail            root.left = toBST(head, mid);            root.right = toBST(mid.next,tail);            return root;        }    }    //双指针遍历    public ListNode getMid(ListNode head,ListNode tail)    {        ListNode slow = head,fast = head;        while(fast!=tail)        {            fast = fast.next;            if(fast!=tail)            {                fast = fast.next;                slow = slow.next;            }        }        return slow;    }}

下面是C++的做法，和上一道题一模一样，直接类似二叉树DFS深度优先遍历即可，不过这里用到了双指针来寻找中间节点

代码如下：

#include <iostream>#include <vector>#include <queue>#include <algorithm>using namespace std;/*struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    TreeNode* sortedListToBST(ListNode* head)    {        return getAll(head,NULL);    }    TreeNode* getAll(ListNode* beg , ListNode* end)    {        if (beg == end)            return NULL;        else        {            ListNode* mid = getMid(beg,end);            TreeNode* root = new TreeNode(mid->val);            root->left = getAll(beg, mid);            root->right = getAll(mid->next, end);            return root;        }    }    ListNode* getMid(ListNode* head, ListNode* end)    {        ListNode* fast = head;        ListNode* slow = head;        while (fast != end)        {            fast = fast->next;            if (fast != end)            {                fast = fast->next;                slow = slow->next;            }        }        return slow;    }};