完全二叉树节点数

问题

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

思考

Examining the structure

A complete binary tree with 10 leaves on its 10th level is being analyzed. The main focus is to determine the maximum number of nodes within this tree.

Clarifying the structure

I'm defining a complete binary tree, noting that while all levels except the last are filled, the last level's nodes are left-aligned. The focus is on the 10th level having 10 leaves.

Determining the final level

I’m piecing together if the 10th level with 10 leaves is the last level or if there could be more nodes below it, considering the structure of the complete binary tree.

Charting node distribution

I'm considering the tree's configuration: the 10th level has 10 nodes, with the first 9 levels fully populated to maximize the total number of nodes.

Confirming tree structure

I’m verifying that the first 9 levels of a complete binary tree are fully populated, totaling 511 nodes, with the 10th level fitting 10 nodes, left-aligned.

Calculating full levels

I’m thinking through the number of nodes from level 1 to level 9, which gives 511 nodes. Turning my attention to the 10th level, which has 10 nodes, I know these nodes are left-aligned.

Confirming leaf nodes

I’m thinking through whether the 10th level has exactly 10 leaf nodes, as typically, nodes on the last level in a binary tree are leaves without children.

Establishing the boundary

The 10th level with 10 leaf nodes is confirmed as the final level, clarifying that these nodes have no children and there are no further levels beyond the 10th.

Mapping node distribution

OK, let me see. The 9th level, with 256 nodes, could have 0, 1, or 2 children each. To maintain completeness, those children must be left-aligned.

Maximizing node allocation

To achieve 10 nodes at the 10th level, it's strategic for 9th-level nodes to have 2 children each, optimizing the use of the 10 available nodes in the fullest and most effective way.

Mapping node distribution

I’m placing 10 child nodes among 256 parents on the 9th level, ensuring the 10th level ends up with exactly 10 nodes. Progress is steady as nodes are distributed left to right.

Evaluating node limits

I’m curious if 521 nodes are the maximum, considering the 10th level has 10 nodes. It’s challenging to envision more than 521 nodes without exceeding the defined tree structure.

Confirming the structure

The tree has a height of 10, with the first 9 levels fully populated and the 10th level containing 10 nodes. The total is 521 nodes.

Thought about 完全二叉树节点数量，持续 43 秒

回答

先明确概念：“完全二叉树”要求除了最后一层外，其余各层都要“完全填满”；并且在最后一层中，节点是从左到右依次排列的。

题意是：这棵完全二叉树的第 10 层有 10 个叶子结点，问整棵树最多有多少个结点？这里通常隐含的是“第 10 层即为该树的最底层”，因为若第 10 层的节点还有孩子，就会出现第 11 层。

分析思路

前 9 层是满的
在完全二叉树中，如果有第 10 层，那么前 9 层必定是“满的”（即每一层都完全填满）。
- 第 1 层（根）：1 个结点
- 第 2 层：2 个结点
- 第 3 层：4 个结点
- …
- 第 9 层：2^8 = 256 个结点
将 1～9 层的节点数相加：
$1 + 2 + 4 + 8 + \cdots + 2^8 \;=\; 2^9 - 1 \;=\; 511$
第 10 层有 10 个节点
由于是完全二叉树，第 10 层的节点从左到右依次“贴紧”，共有 10 个节点。这些节点都是叶子（没有第 11 层）。
整棵树的总节点数
将前 9 层的 511 个节点，加上第 10 层的 10 个节点，总共得到
$511 + 10 = 521$

这样摆放就能保证第 9 层是“满”的，第 10 层有 10 个（并且都靠左排列），符合完全二叉树的定义，且第 10 层确为叶子。此时整棵树的节点数也已经达到最大可能。

答案

这棵完全二叉树最多有 521 个节点。

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