LeetCode - Convert Sorted List to Binary Search Tree
Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example 1: Input: head = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST. Example 2: Input: head = [] Output: [] Example 3: Input: head = [0] Output: [0] Example 4: Input: head = [1,3] Output: [3,1] Constraints: The number of nodes in head is in the range [0, 2 * 104]. -10^5 <= Node.val <= 10^5
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode sortedListToBST(ListNode head) { if (head == null) { return null; } if (head.next == null) { return new TreeNode(head.val); } ListNode slow = head; ListNode fast = head.next.next; while(fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } ListNode tmp = slow.next; slow.next = null; TreeNode node = new TreeNode(tmp.val); node.left = sortedListToBST(head); node.right = sortedListToBST(tmp.next); return node; } }
这种接法巧妙的点是如何找到中点,和递归的时候如何切断中点与前后两个linkedlist的链接的。
posted on 2020-10-13 12:50 IncredibleThings 阅读(109) 评论(0) 收藏 举报