题目描述

给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

注意：

• 有序链表构建的高度平衡的二叉树就是平衡二叉树
• 简单容易想到的就是链表转为有序数组，直接构建，但是这样需要额外的空间。
• 要求空间O(1)的情况下可以使用双指针快慢指针，当快指针走到区间的尽头，满慢指针刚好走到 中间

解答

链表转有序集合，直接递归构建

/**
 * 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 {
    List<Integer> list = new ArrayList<>();

    public TreeNode sortedListToBST(ListNode head) {
        if(head == null){
            return null;
        }
        while(head != null){
            list.add(head.val);
            head = head.next;
        }
        int len = list.size();

       
        return buildBST(0,len-1);
        
        
    }
    public TreeNode buildBST(int start,int end){
        if(start > end ){
            return null;
        }
        
        int mid = (start + end) / 2;

        TreeNode node = new TreeNode(list.get(mid));

        
        node.left = buildBST(start,mid -1);
        node.right = buildBST(mid + 1,end);
        return node;
    }
}

快慢指针

/**
 * 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;
        }
        return helper(head,null);
    }
    private TreeNode helper(ListNode head,ListNode tail){
        if(head == tail){
            return null;
        }
        // 快慢指针
        ListNode fast = head;
        ListNode slow = head;
        // 慢指针 定位区间中间
        while(fast != tail && fast.next != tail){
            fast = fast.next.next;
            slow = slow.next;
        }
        TreeNode root = new TreeNode(slow.val);
        root.left = helper(head,slow);
        root.right = helper(slow.next,tail);
        return root;
    }
}