LeetCode 109. Convert Sorted List to Binary Search Tree

原题链接在这里：https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

题目：

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

题解：

找到List中点当BST的root. 中点之前当左子树，中点之后当右子树。

递归调用，终止条件有两个: 第一，当ListNode 为 null 时返回null.

第二，当只有一个ListNode时返回由当前点生成的TreeNode.

Note：此处findMidLeft找到的是中点的前一个点，然后拆成三段，中点前一段，中点单独一个，中点后一段.

Time Complexity: O(n*logn), 因为T(n) = 2*T(n/2) + n, findMidLeft 用了n时间, n是List长度。

Space: O(logn),  一共用了O(logn) 层stack.

AC Java:

/**
 * 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; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head == null){
            return null;
        }
        
        if(head.next == null){
            return new TreeNode(head.val);
        }
        ListNode midLeft = findMidLeft(head);
        ListNode mid = midLeft.next;
        ListNode midRight = mid.next;
        
        //Break
        midLeft.next = null;
        mid.next = null;
        
        //Create root;
        TreeNode root = new TreeNode(mid.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(midRight);
        return root;
    }
    
    private ListNode findMidLeft(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        ListNode walker = head;
        ListNode runner = head;
        while(runner.next != null && runner.next.next != null && runner.next.next.next != null){
            walker = walker.next;
            runner = runner.next.next;
        }
        return walker;
    }
    
}

类似Convert Sorted Array to Binary Search Tree.