109. Convert Sorted List to Binary Search Tree

Question

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

Thinking:

• Method 1:递归， 双指针
  • 这道题的思想其实是和108. Convert Sorted Array to Binary Search Tree想法是一致的。
  • 通过双指针的找到链表的中点。
  • 递归将结点一个个赋值到二叉查找树中。

/**
 * 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;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        ListNode pre = null;
        while(fast != null && fast.next != null){
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        TreeNode ret = new TreeNode(slow.val);
        ret.left = sortedListToBST(dummy.next);
        ret.right = sortedListToBST(slow.next);
        return ret;
    }
}

二刷

有一个疑惑，当node.left = sortedListToBST(dummy.next);中的dummy.next改成head时就会爆栈。

/**
 * 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;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = null;
        ListNode slow = dummy;
        ListNode fast = dummy;
        while(fast != null && fast.next != null){
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        TreeNode node = new TreeNode(slow.val);
        node.left = sortedListToBST(dummy.next);
        node.right = sortedListToBST(slow.next);
        return node;
    }
}