【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:
public TreeNode sortLisTreeNodeRecur(int size){
if(size <= 0)
return null; TreeNode left = sortLisTreeNodeRecur(size/2);
TreeNode root = new TreeNode(current.val);
current = current.next;
TreeNode right = sortLisTreeNodeRecur(size - size/2 - 1); root.left = left;
root.right = right; return root;
}
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:
/**
* 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; }
* }
*/
public class Solution {
private ListNode current;
public int getLength(ListNode head){
int answer = 0;
while(head != null){
answer++;
head = head.next;
} return answer;
}
public TreeNode sortLisTreeNodeRecur(int size){
if(size <= 0)
return null; TreeNode left = sortLisTreeNodeRecur(size/2);
TreeNode root = new TreeNode(current.val);
current = current.next;
TreeNode right = sortLisTreeNodeRecur(size - size/2 - 1); root.left = left;
root.right = right; return root;
}
public TreeNode sortedListToBST(ListNode head) {
current = head;
int size = getLength(head);
return sortLisTreeNodeRecur(size);
}
}
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