就叶子数而言，完整 k 叉树中的节点总数是多少? [英] What is the total number of nodes in a full k-ary tree, in terms of the number of leaves?

问题描述

我正在做一种独特形式的 Huffman 编码，并且正在构建一个完整的 k-ary(在这种特殊情况下，3-ary)树(每个节点将有 0 或 k 个孩子)，我知道有多少在我建造它之前它就会有.如何根据叶子数计算树中的节点总数?

I am doing a unique form of Huffman encoding, and am constructing a k-ary (in this particular case, 3-ary) tree that is full (every node will have 0 or k children), and I know how many leaves it will have before I construct it. How do I calculate the total number of nodes in the tree in terms of the number of leaves?

我知道在完全二叉树(2-ary)的情况下，这个公式是 2L - 1，其中 L 是叶子的数量.我想将此原则扩展到 k 叉树的情况.

I know that in the case of a full binary tree (2-ary), the formula for this is 2L - 1, where L is the number of leaves. I would like to extend this principle to the case of a k-ary tree.

推荐答案

想想如何证明一棵完整二叉树的结果，你会看到一般的做法.对于全二叉树来说，高度h，节点数N是

Think about how to prove the result for a full binary tree, and you'll see how to do it in general. For the full binary tree, say of height h, the number of nodes N is

N = 2^{h+1} - 1

为什么?因为第一层有2^0个节点，第二层有2^1个节点，一般来说，k层有2^{k-1} 个节点.将这些加起来总共 h+1 个级别(所以高度 h)给出

Why? Because the first level has 2^0 nodes, the second level has 2^1 nodes, and, in general, the kth level has 2^{k-1} nodes. Adding these up for a total of h+1 levels (so height h) gives

N = 1 + 2 + 2^2 + 2^3 + ... + 2^h = (2^{h+1} - 1) / (2 - 1) = 2^{h+1} - 1

叶子总数L就是最后一层的节点数，所以L = 2^h.因此，通过替换，我们得到

The total number of leaves L is just the number of nodes at the last level, so L = 2^h. Therefore, by substitution, we get

N = 2*L - 1

对于 k-ary 树，除了 2 没有任何变化.所以

For a k-ary tree, nothing changes but the 2. So

N = 1 + k + k^2 + k^3 + ... + k^h = (k^{h+1} - 1) / (k - 1)

L = k^h

所以一点代数可以带你完成最后一步

and so a bit of algebra can take you the final step to get

N = (k*L - 1) / (k-1)

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