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完全二叉树节点数

創建於：2025年1月20日

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問題

有一棵完全二叉树，它的第10层只有10个叶子结点，那么这棵树最多有多少个节点？

思考

Assessing the tree's structure

A complete binary tree's node count can be calculated by doubling the number of nodes in previous layers, ultimately summing these values.

Crunching node counts

I’m working through the number of nodes on the 9th level by doubling the previous level's count. This leads me to explore the potential node count for a fully complete 10th level.

Maximizing node count

Mapping out the distribution of 10 nodes on the 10th level, I count the possible children of the first 10 nodes from the 9th level, aiming to maximize the total tree size.

Clarifying the definition

I’m ensuring the structure adheres to the definition of a complete binary tree, where the last level has 10 left-aligned nodes, and the previous level (9th) has 256 nodes filled.

Distributing children

OK, let me see. I’m distributing 10 children across the 9th level's 5 nodes, considering the left-aligned requirement of the 10th level in a complete binary tree.

Allocating children

Working through how to distribute 10 children across the 9th level's parents, ensuring they are placed from left to right to adhere to the complete binary tree's structure.

Distributing children

OK, let me see. I’m placing 10 children across 9th level parents, ensuring each parent gets children from left to right, adhering to the complete binary tree structure.

Mapping out children placement

I’m ensuring each parent gets two children until all are distributed, adhering to the complete binary tree rules and concluding the 10 children setup aptly.

Confirming distribution

Balancing the distribution, I’m verifying the first 5 parents in the 9th level each have 2 children, ensuring the 10th level matches the complete binary tree structure.

Assessing distribution possibilities

I’m examining if the 10th level could partially extend to the 11th level. Given the 10 leaf nodes, they align well with the complete binary tree structure.

Ensuring completeness

I'm focusing on verifying that the 10th level, with 10 leaf nodes, is the final level. This setup maintains the complete binary tree structure.

Ensuring tree structure

OK, I’ve confirmed the 9th level is fully populated with 256 nodes. The 10th level, featuring 10 leaf nodes, completes the tree's structure, totaling 521 nodes.

Thought about 完全二叉树节点数，持续 59 秒

回答

要想求得这棵完全二叉树的最大结点数，关键在于理解“完全二叉树”以及“第10层只有10个叶子结点”对整棵树结构的限制。下面分步骤说明思路：

1. 什么是完全二叉树

完全二叉树指的是：

  • 除了最末层之外，其他各层都被完全填满；
  • 最末层的结点都尽可能地靠左排列。

因此，如果某一层（比如第10层）部分填充了结点，则它之前的所有层(1～9层)必须是“满”的。

2. 前 9 层的结点数

一个满二叉树(层数为 kk) 的结点总数是

2k−1.2^k - 1.
  • 第 1 层有 20=12^0 = 1 个结点。
  • 第 2 层有 21=22^1 = 2 个结点。
  • …
  • 第 9 层有 28=2562^8 = 256 个结点。

因此，前 9 层的结点总数为

29−1=511.2^9 - 1 = 511.

3. 第 10 层只有 10 个叶子结点

题目说明“第10层只有10个叶子结点”，并且这些叶子结点是“完全二叉树”意义下最末层的结点，说明没有第11层。

  • 由于是完全二叉树，第 10 层的这 10 个结点必须从左到右依次占据第 10 层的前 10 个位置。
  • 能有第 10 层，意味着第 9 层已经全部填满(即 256 个结点全部存在)，然后根据完全二叉树的特性，为最左边的若干第 9 层父节点依次填上它们的左右孩子，直到在第10层放满这10个叶子。

要放 10 个结点，只需前 5 个父节点各拥有左右孩子即可（每个父节点 2 个孩子，合计 10 个），这样恰好满足“从左到右连续的 10 个叶子”，其他第9层的父节点不再有孩子。

4. 全树最大结点数

前 9 层结点数为 511511，第 10 层有 1010 个叶子结点，所以整棵树的结点总数为

511+10=521.511 + 10 = 521.

这就是在“第 10 层只有 10 个叶子结点”前提下，整棵完全二叉树所能拥有的最大结点数。

答案：521

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