Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. 这道题让我们求二叉搜索树的某个节点的中序后继节点,那么我们根据BST的性质知道其中序遍历的结果是有序的, 是我最先用的方法是用迭代的中序遍历方法,然后用
Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Example 1: Input: root = [2,1,3], p = 1 Output: 2 Explanation: 1's in-or
Add an Arbiter to Replica Set 在集群中加入仲裁节点,当集群中主节点挂掉后负责选出新的主节点,仲裁节点也是一个mongo实力,但是它不存储数据. 1.仲裁节点消耗很小的资源,而且不需要专用的服务器. 2.不能把仲裁节点安装到集群中的其它节点服务器上. 3.journal.enabled to false 减少资源占用. 4.mallFiles to true 减少资源占用. 注意上面3,4的配置不用设置到其它集群节点中. 5.rs.addArb("m1.exampl
[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
二叉搜索树是常用的概念,它的定义如下: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search