根据有序链表构造平衡的二叉查找树

Posted 2021-12-28 zhuge134

leetcode地址：

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

难度：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


解题思路：

分为两步：

1. 把链表转存到一个数组中，问题转化为：将一个有序数组转化为一个平衡二叉查找树。

2. 取数组的中点为根节点，那么根节点的左子树是由左半边的数组生成的，右子树是由数组右半边生成的，分别对数组左半边和右半边进行递归调用

 

代码：

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;