Description

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:

BST

Solution

DFS

首先找到list的中间节点，然后递归调用即可。递归的时候，要么把链表拆分成三部分，要么把tail节点传入，作为递归的终止条件。

把tail节点传入的写法更加简洁和直观，要记住哦！

/**
 * 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; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
            return null;
        }
        
        return toBSTRecur(head, null);
    }
    
    // construct a BST using nodes between [head, tail)
    public TreeNode toBSTRecur(ListNode head, ListNode tail) {
        if (head == tail) {
            return null;
        }
        
        ListNode slow = head;
        ListNode fast = head;
        
        while (fast != tail && fast.next != tail) {
            fast = fast.next.next;
            slow = slow.next;
        }
        
        TreeNode root = new TreeNode(slow.val);
        root.left = toBSTRecur(head, slow);
        root.right = toBSTRecur(slow.next, tail);
        
        return root;
    }
}