LeetCode OJ 109. Convert Sorted List to Binary Search Tree

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

涉及到二叉树的问题用递归的方法很容易理解。这个问题要求把一个升序排序的链表转换为一颗height balanced BST。转换的方式就是把链表的中间值作为树的根节点，中间值的左半部分转换为二叉树的左子树，中间值的右半部分转换为二叉树的右子树。

递归解决问题的步骤如下：

1. 如果链表的长度为0，返回null；

2. 如果链表长度为1，创建新的二叉树节点node，node.left=null，node.right=null，node.val=head.val。

3. 否则的话找到链表的中间节点，即第mid = length%2==0?length/2:length/2+1个节点为根节点node。此时链表被分成两个部分，前半部分的长度为mid-1，后半部分的长度为length-mid，递归求这两部分的结果，并把他们分别赋值给根节点的左子树和右子树。代码如下：

public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        ListNode pointer = head;
                
        int length = 0;
        while(pointer!=null){
            pointer = pointer.next;
            length++;
        }
        
        return listToBST(head, length);
 
    }
    
    public TreeNode listToBST(ListNode head,int length){
        if(length==0) return null;
        if(length==1){
            TreeNode node = new TreeNode(head.val);
            node.left = null;
            node.right = null;
            return node;
        }
        if(length==2){
            TreeNode node = new TreeNode(head.val);
            node.left = null;
            node.right = listToBST(head.next, 1);
            return node;
        }
        int mid = length%2==0?length/2:length/2+1;
        
        ListNode pointer = head;
        ListNode righthead = null;
        int i = 1;
        while(pointer.next!=null && i<mid-1){
            pointer = pointer.next;
            i++;
        }
        TreeNode node = new TreeNode(pointer.next.val);
        righthead = pointer.next.next;
        pointer.next = null;
        node.left = listToBST(head, mid-1);
        node.right = listToBST(righthead, length-mid);
        
        return node;
    }
}