*[Lintcode]Convert Sorted List to Balanced BST-优快云博客

本文介绍了一种将升序单链表转换为高度平衡二叉搜索树的方法。通过递归方式从中间节点开始构建，确保左右子树的高度尽可能相同。

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

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Example

               2
1->2->3  =>   / \
             1   3

二叉树中序遍历后可得到递增序列，同理，由递增序列可得到二叉树。这里采用从下向上计算的方法构建二叉树，这样可以利用linked list有序这一特征。

public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    private ListNode curr = null;
     
    public TreeNode sortedListToBST(ListNode head) {  
        int length = getLength(head);
        curr = head;
        return helper(0, length - 1);
    }
    
    TreeNode helper(int start, int end) {
        if(start > end) return null;
        
        int mid = (start + end) / 2;
        TreeNode left = helper(start, mid - 1);
        TreeNode root = new TreeNode(curr.val);
        curr = curr.next;
        TreeNode right = helper(mid + 1, end);
        
        root.left = left;
        root.right = right;
        return root;
    }
    
    int getLength(ListNode head) {
        int res = 0;
        while(head != null) {
            res ++;
            head = head.next;
        }
        return res;
    }
}