什么是没有。具有 n 个标记节点的不同二叉树的可能性？
What is the no. of distinct binary trees possible with n labeled nodes?
在此 link 中给出：
For a binary tree with n nodes, the no. of edges is n−1. So, this
problem can be reduced to the no. of ways in which we can make n−1
edges from n vertices. An edge can be made either as a left child of
a node or as a right child. Hence, for n nodes, we have 2n
possibilities for the first edge, 2n−1 for the second edge and so on.
Thus, for n−1 edges, the total no. of ways
= 2n×(2n−1)×(2n−2)×….×(2n–(n–2))
= 2n×(2n−1)×(2n−2)×….×(n+2)
=(2n)!/(n+1)!
我确实知道第一条边可以有 2n 种可能性，因为对于每个节点都有左右子选项。我想不通第二条边怎么会有 2n-1 种可能性？
n=3时第二条边的可能性是多少？
how second edge can have 2n-1 possibilities?
在您添加第一条边之前，有 2n 种可能性。
添加第一个边后，一个可能性被占用，只剩下2n-1个可能性。
在第二条边之后只剩下 2n-2 种可能，依此类推
对于 n=3，有 6!/4!=30 个变体。只需检查：有 5 个配置，每个配置有 6 个排列：
/\ / / \ \
/ \ / \
在此 link 中给出：
For a binary tree with n nodes, the no. of edges is n−1. So, this problem can be reduced to the no. of ways in which we can make n−1 edges from n vertices. An edge can be made either as a left child of a node or as a right child. Hence, for n nodes, we have 2n possibilities for the first edge, 2n−1 for the second edge and so on. Thus, for n−1 edges, the total no. of ways
= 2n×(2n−1)×(2n−2)×….×(2n–(n–2))
= 2n×(2n−1)×(2n−2)×….×(n+2)
=(2n)!/(n+1)!
我确实知道第一条边可以有 2n 种可能性，因为对于每个节点都有左右子选项。我想不通第二条边怎么会有 2n-1 种可能性？
n=3时第二条边的可能性是多少？
how second edge can have 2n-1 possibilities?
在您添加第一条边之前，有 2n 种可能性。
添加第一个边后，一个可能性被占用，只剩下2n-1个可能性。
在第二条边之后只剩下 2n-2 种可能，依此类推
对于 n=3，有 6!/4!=30 个变体。只需检查：有 5 个配置，每个配置有 6 个排列：
/\ / / \ \
/ \ / \