和上题思路基本一致,不同的地方在于,链表不能随机访问中间元素。

int listLength(ListNode* node)
{
int n = ;
while (node)
{
n++;
node = node->next;
}
return n;
}
ListNode* nth_node(ListNode* node, int n)
{
while (--n)node = node->next;
return node;
}
TreeNode* sortedListToBST(ListNode* head)
{
sortedListToBST(head, listLength(head));
}
TreeNode* sortedListToBST(ListNode* head, int len)
{
if (len == )return nullptr;
if (len == )return new TreeNode(head->val); TreeNode* root = new TreeNode(nth_node(head, len / + )->val);
root->left = sortedListToBST(head, len / );
root->right = sortedListToBST(head, len / + ); return root;
}

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