[抄题]: Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST. Basically, the deletion can be divided into two stages: Search for a node to remove. If
[试题描述]定义一个函数,输入一个链表,删除无序链表中重复的节点 [参考代码] 方法一: Without a buffer, we can iterate with two pointers: "current" does a normal iteration, while "runner" iterates through all prior nodes to check for dups Runner will only see one dup per node