题目描述:
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
思路:二叉查找树的中序遍历就是排序的数,那么我们可以通过中序遍历来构建这个树
直接看代码,下面一般是改进的 ,这里题目要求的是,当节点是偶数的时候,以n/2+1个作为中间节点。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; next = null; }
* }
*/
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
//二叉查找树的中序遍历就是排序的数,那么我们可以通过中序遍历来构建这个树
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
if(head==null)
return null;
if(head.next==null)
return new TreeNode(head.val);
//至少有两个节点才能用快慢节点切割,不然节点为1的时候 会出现分为 1,0这样的死循环
ListNode fast=head;
ListNode slow=head;
ListNode pre=null;
while(fast!=null&&fast.next!=null){
pre=slow;
slow=slow.next;
fast=fast.next.next;
}
if(pre!=null)
pre.next=null;
ListNode mid=slow;//慢指针指向的就是中间节点
ListNode nextHead=slow.next;
TreeNode node=new TreeNode(mid.val);
if(pre==null)
node.left=sortedListToBST(null);
else
node.left=sortedListToBST(head);
node.right=sortedListToBST(nextHead);
return node;
}
//代码改进
public TreeNode sortedListToBST(ListNode head) {
return sortedListToBST(head,null);
}
private TreeNode sortedListToBST(ListNode head,ListNode end){
if(head==end)
return null;
ListNode fast=head;
ListNode slow=head;
while(fast!=end&&fast.next!=end){
fast=fast.next.next;
slow=slow.next;
}
TreeNode root=new TreeNode(slow.val);
root.left=sortedListToBST(head,slow);
root.right=sortedListToBST(slow.next,end);
return root;
}
}