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【leetcode刷题笔记】Convert Sorted List to Binary Search Tree

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


 

题解:开始想到的方法比较偷懒,直接遍历一遍数组,把每个ListNode对应的值放到数组里面,然后把数组转换成BST,想来这个题的本意肯定不是这样。

自己也想到了bottom-up的方法,但是没想明白怎么个bottom-up法,真的是昨天做梦都在想=。=

今天早上终于弄懂了,用递归,不过这个递归比较难理解。

递归函数的定义是这样的 public TreeNode sortLisTreeNodeRecur(int size) ,参数size的意义可以理解为要建立的树的节点个数。用一个current变量作为linked list的游标。递归函数的实现如下:

 1 public TreeNode sortLisTreeNodeRecur(int size){ 2         if(size <= 0) 3             return null; 4          5         TreeNode left = sortLisTreeNodeRecur(size/2); 6         TreeNode root = new TreeNode(current.val); 7         current = current.next; 8         TreeNode right = sortLisTreeNodeRecur(size - size/2 - 1); 9         10         root.left = left;11         root.right = right;12         13         return root;14     }

可以看到,它先递归调用自己建立整个左子树,在这个过程中,current作为全局私有变量,在左子树建立完成以后,就自动移动到根节点对应的ListNode处了,所以建立完左子树后建立根节点,后移current,然后递归的建立整个右子树。

以Linked List = 1->2->3->4->5来画图说明:

代码如下:

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { val = x; next = null; } 7  * } 8  */ 9 /**10  * Definition for binary tree11  * public class TreeNode {12  *     int val;13  *     TreeNode left;14  *     TreeNode right;15  *     TreeNode(int x) { val = x; }16  * }17  */18 public class Solution {19     private ListNode current;20     public int getLength(ListNode head){21         int answer = 0;22         while(head != null){23             answer++;24             head = head.next;25         }26         27         return answer;28     }29     public TreeNode sortLisTreeNodeRecur(int size){30         if(size <= 0)31             return null;32         33         TreeNode left = sortLisTreeNodeRecur(size/2);34         TreeNode root = new TreeNode(current.val);35         current = current.next;36         TreeNode right = sortLisTreeNodeRecur(size - size/2 - 1);37         38         root.left = left;39         root.right = right;40         41         return root;42     }43     public TreeNode sortedListToBST(ListNode head) {44         current = head;45         int size = getLength(head);46         return sortLisTreeNodeRecur(size);47     }48 }