【原创】leetCodeOj ---Convert Sorted List to Binary Search Tree 解题报告

原题地址：

https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

 

题目内容：

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

 

方法：

单纯。如何安排插入顺序，使得一棵二叉排序树接近平衡？

你从中间开始插嘛。这样左子树和右子树之差或者为0，或者为1。

所以，这个问题的本质是找链表的中间节点。

找到中间节点后，递归产生左子树和右子树，在找左边子链表的中间节点和右边子链表的中间节点即可。

当然啦，最后返回中间节点。

 

全部代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        int len = countLen(head);
        if (len == 0)
            return null;
        return constructTree(head,len); // len is the number of nodes in list.
    }
    
    private TreeNode constructTree(ListNode head,int len)
    {
        if (head == null || len <= 0)
            return null;
        int mid = len / 2 + 1;
        TreeNode tmp = new TreeNode(head.val);
        tmp.left = null;
        tmp.right = null;
        if (mid == 0)
            return tmp;
        ListNode tar = findMidNode(head,mid);
        tmp.val = tar.val;
        tmp.left = constructTree(head,mid - 1);
        tmp.right = constructTree(tar.next,len - mid);
        return tmp;
    }
    
    private ListNode findMidNode(ListNode head,int mid)
    {
        int count = 1;
        while (mid != count++)
            head = head.next;
        return head;
    }
    
    private int countLen(ListNode head)
    {
        int len = 0;
        while (head != null)
        {
            len ++;
            head = head.next;
        }
        return len;
    }
}