根據有序鏈表構造平衡的二叉查找樹

leetcode地址：

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/

難度：中等

 

描述：

Given 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:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5


解題思路：

分爲兩步：

1. 把鏈表轉存到一個數組中，問題轉化爲：將一個有序數組轉化爲一個平衡二叉查找樹。

2. 取數組的中點爲根節點，那麼根節點的左子樹是由左半邊的數組生成的，右子樹是由數組右半邊生成的，分別對數組左半邊和右半邊進行遞歸調用

 

代碼：

public class SortedListToBST {    public TreeNode sortedListToBST(ListNode head) {        int size = 0;        ListNode p = head;        while (p != null) {            size++;            p = p.next;        }        ListNode[] listNodes = new ListNode[size];        p = head;        int index = 0;        while (p != null) {            listNodes[index++] = p;            p = p.next;        }        return sortedListToBST(listNodes, 0, listNodes.length);    }    public TreeNode sortedListToBST(ListNode[] listNodes, int start, int end) {        if (start >= end) {            return null;        }        int mid = (end + start) / 2;        TreeNode root = new TreeNode(listNodes[mid].val);        root.left = sortedListToBST(listNodes, start, mid);        root.right = sortedListToBST(listNodes, mid + 1, end);        return root;    }}