V-DIMENSION
VDS ECOLOGY DATA UPDATE
BLOCK HEIGHT (0-1,489,472)
YPOO
Current Resonance Level(BTC/Vollar) 1:10,123
Resonated Vollar Price $3.78
Current VID Count 5,238,418
Super Master Nodes(Network-wide) 9,569
Super Master Node Revenue 1.75
Miner Revenue 1.75
Nathony u denal oal ibus ha froer
VDS ECOLOGY DATA UPDATE
BLOCK HEIGHT (0-1,489,472)
YPOO
Current Resonance Level(BTC/Vollar) 1:10,123
Resonated Vollar Price $3.78
Current VID Count 5,238,418
Super Master Nodes(Network-wide) 9,569
Super Master Node Revenue 1.75
Miner Revenue 1.75
Nathony u denal oal ibus ha froer
V-DIMENSION
VDS ECOLOGY DATA UPDATE
BLOCK HEIGHT(0-1,485,451
WPOOL
Current Resonance Level(BTC/Vollar) 1:10,109
Resonated Vollar Price $4.12
Current VID Count 5,238,375
Super Master Nodes(Network-wide) 9,570
Super Master Node Revenue 1.75
Miner Revenue 1.75
Nabheg a doudl on ila ha foror.
VDS ECOLOGY DATA UPDATE
BLOCK HEIGHT(0-1,485,451
WPOOL
Current Resonance Level(BTC/Vollar) 1:10,109
Resonated Vollar Price $4.12
Current VID Count 5,238,375
Super Master Nodes(Network-wide) 9,570
Super Master Node Revenue 1.75
Miner Revenue 1.75
Nabheg a doudl on ila ha foror.
543. Diameter of Binary Tree
Given the root of a binary tree, return 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.
The length of a path between two nodes is represented by the number of edges between them.
解题重点:
1. 经过某个节点路径的长度等于节点左右子树的深度之和减去1
2. 二叉树的直径就是所有节点的路径长度中的最大值
3. 先递归调用左儿子和右儿子求得它们为根的子树的深度 LL 和 RR ,则该节点为根的子树的深度即为max(L,R)+1
4. 节点的长度等于L+R+1
5. 递归搜索每个节点并设一个全局变量 ans记录最大值
Given the root of a binary tree, return 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.
The length of a path between two nodes is represented by the number of edges between them.
解题重点:
1. 经过某个节点路径的长度等于节点左右子树的深度之和减去1
2. 二叉树的直径就是所有节点的路径长度中的最大值
3. 先递归调用左儿子和右儿子求得它们为根的子树的深度 LL 和 RR ,则该节点为根的子树的深度即为max(L,R)+1
4. 节点的长度等于L+R+1
5. 递归搜索每个节点并设一个全局变量 ans记录最大值
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