LeetCode-109. Convert Sorted List to Binary Search Tree

LeetCode-109. Convert Sorted List to Binary Search Tree

Posted by Jae on September 16, 2019

1、题目

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

2、思路

这道题让我们将一个有序的单链表转成一个二叉平衡树,二叉平衡树的定义是一颗空树或者左右子树的高度差不大于1的二叉树,这里已经给了单链表是有序,所以只需要每次找到 链表的中点作为二叉平衡树的树根,然后递归的将中点左边的链表转成根节点的左子树,中点右边的链表转成根节点的右子树即可, 如何找到单链表的中心节点,可以使用快慢指针。

3 实现

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
}


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;
        if (head.next == null) return new TreeNode(head.val);

        ListNode fast = head, slow = head;
        ListNode prev = new ListNode(0);
        prev.next = head;
        while (fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            prev = prev.next;
        }
        prev.next = null;
        TreeNode root = new TreeNode(slow.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(slow.next);

        return root;
    }
}