【Leetcode】Convert Sorted List to Binary Search Tree (BST)

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

这道题采用的基本思路和数组构建BST差不多，但是有几点不同

1. 你需要统计链表的长度

2. 你无法随机访问链表，所以为了避免多次访问链表所提高的时间复杂度，我们决定顺序访问

3. 顺序访问就要求在构建二叉树的时候不能采用前序遍历，而是中序遍历

即先构造左节点，构造根节点，连接根和左节点，更新根，连接根和右节点

代码如下

public TreeNode sortedListToBST(ListNode head) {
if (head == null)
return null;
ListNode runner = head;
int count = 0;
while (runner != null) {
runner = runner.next;
count++;
}
ArrayList<ListNode> nodeList = new ArrayList<ListNode>();
nodeList.add(head);
return helper(nodeList, 0, count - 1);
}

public TreeNode helper(ArrayList<ListNode> nodeList, int left, int right) {
if (left > right)
return null;
int middle = (left + right) / 2;
TreeNode leftNode = helper(nodeList, left, middle - 1);
TreeNode root = new TreeNode(nodeList.get(0).val);
root.left = leftNode;
nodeList.set(0, nodeList.get(0).next);
root.right = helper(nodeList, middle + 1, right);
return root;
}
可以思考一下为什么nodeList.set(0, nodeList.get(0).next)这一行代码一定要在root.left=leftNode之后, root.right=helper(nodeList,
middle+1, right)之前