109. Convert Sorted List to Binary Search Tree

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

从上往下递归构造二叉树，这样每次都要遍历时间复杂度高一些

1ms

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *//** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode sortedListToBST(ListNode head) {        if(head==null) return null;        ListNode tail=null;        return toBST(head,tail);        }    public TreeNode toBST(ListNode head,ListNode tail){        ListNode fast=head;        ListNode slow=head;        if(head==tail) return null;        while(fast!=tail&&fast.next!=tail){            fast=fast.next.next;            slow=slow.next;        }        TreeNode newhead=new TreeNode(slow.val);        newhead.left=toBST(head,slow);        newhead.right=toBST(slow.next,tail);        return newhead;    }}

为什么这个更慢呢，。。

使用108题处理数组的方法，先把列表的值存到一个list里然后递归不断调用

3ms

public class Solution {    public TreeNode sortedListToBST(ListNode head) {        if(head==null) return null;        List<ListNode> list=new ArrayList<ListNode>();        while(head!=null){            list.add(head);            head=head.next;        }        return sort(list,0,list.size()-1);        }    public TreeNode sort(  List<ListNode> list,int start,int end){        if(start>end) return null;        int mid=start+(end-start)/2;        TreeNode root=new TreeNode(list.get(mid).val);        root.left=sort(list,start,mid-1);        root.right=sort(list,mid+1,end);        return root ;    }}

使用了中序遍历的顺序来构建这棵树。不用每次都遍历链表。而且不用另开空间

是在superchao在discuss中记录下来的，用java改写了一下第一次提交竟然说error。。。过了一会提交就过了最近这个leetcode改了之后时常出现这种问题尴尬

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *//** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    ListNode list;    public TreeNode sortedListToBST(ListNode head) {        this.list=head;        return sort(count(head));    }    public int count(ListNode head){        int size=0;        while(head!=null){            size++;            head=head.next;        }        return size;    }    public TreeNode sort(int n){        if(n==0)            return null;        TreeNode node=new TreeNode(0);        node.left=sort(n/2);  //节点个数        node.val=list.val;   //前n/2是左子树的节点，这个节点刚好是 根节点        list=list.next;        node.right=sort(n-n/2-1); //这是右子树的节点           return node;    }}

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