Leetcode109——Convert Sorted List to Binary Search Tree
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文章作者:Tyan
博客:noahsnail.com | CSDN | 简书
1. 问题描述
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
2. 求解
这个题主要是根据一个有序链表构造二叉查找树(树的左结点小于根节点,根节点小于右结点,子树具有同样的性质)。与有序数组最大的不同在于有序链表只能从前往后遍历,不能像有序数组一样访问任意位置的元素。因此构造时需要按顺序构造,其实有序链表是二叉查找树的中序遍历。因此需要按照中序遍历的顺序进行构建,先构建左子树,再构造根节点,最后构造右子树。由于是链表,每次构造之后头结点应该进行移动,Java中用了一个静态变量来保存根节点的位置。构造方法主要是递归,每次构建子树时都需要将数组分成左右两半,左边的构建左子树,右边的构建右子树,中间元素构造根节点。
/** * 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 { static ListNode h; public TreeNode sortedListToBST(ListNode head) { if(head == null) { return null; } int length = 0; h = head; //得到链表长度 while(head != null) { length++; head = head.next; } return buildBinarySearchTree(h, 0, length - 1); } public TreeNode buildBinarySearchTree(ListNode head, int start, int end) { if(start > end) { return null; } int mid = (start + end) / 2; //先构建左子树 TreeNode left = buildBinarySearchTree(h, start, mid - 1); //再构造根节点 TreeNode root = new TreeNode(h.val); h = h.next; //最后构造右子树 TreeNode right = buildBinarySearchTree(h, mid + 1, end); root.left = left; root.right = right; return root; }}
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