diameter

1. 创建和发送Request消息: 1) 产生一个Request消息时,必须遵守下列规则: · 设置头部的Command code: · 设置头部的 'R' 位: · 设置头部的End-to-End 为本地的唯一值: · Origin-Host 和 Origin-Realm AVPs 必须携带, 用来标识消息的源地址: · Destination-Host 和 Destination-Realm AVPs 需根据以下规则设置: a) 不能被Proxy的消息一定不能带Destination-Re

Diameter是计算机网络中使用的一个认证.授权和审计协议.它从功能更少的RADIUS协议进化而来,并且取代之. Dimater应用通过添加新的命令或属性扩展基础协议,例如使用扩展认证 协议(EAP). 与RADIUS协议比较 名字是一个单词游戏(从RADIUS,半径到DIAMETER,直径),从前任RADIUS协议演化而来(直径是半径的二倍).Diameter不直接后向兼容,但是提供RADIUS升级路径.DIAMETER提供的主要特征(RADIUS中缺乏的)有: * 可靠传输协议(TCP或S

/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public int DiameterOfBinaryTree(TreeNode root)

题目: Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root. Example:Given a bina

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longestpath between any two nodes in a tree. This path may or may not pass through the root. Example:Given a binary tr

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longestpath between any two nodes in a tree. This path may or may not pass through the root. Example:Given a binary tr

The diameter of a tree (sometimes called the width) is the number of nodes on the longest path between two leaves in the tree. The diameter of a tree T is the largest of the following quantities: * the diameter of T’s left subtree* the diameter of T’

The diameter of a tree (sometimes called the width) is the number of nodes on the longest path between two leaves in the tree. The diagram below shows two trees each with diameter nine, the leaves that form the ends of a longest path are shaded (note

题目中的直径定义为: 任意两个节点的最远距离 没想出来,看的答案 思路是:diameter = max(左子树diameter,右子树diameter,(左子树深度+右子树深度+1)) 遍历并更新结果 int res = 1; public int diameterOfBinaryTree(TreeNode root) { helper(root); return res-1; } public int[] helper(TreeNode root) { //两个量分别是节点深度,节点最大dia