完全二叉树节点数

Question

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？

Thought

Breaking down the problem

I’m analyzing a complete binary tree where the 10th level contains 10 leaf nodes. Unlike typical trees, the 10th level may be the last, making it crucial to understand the specific tree structure.

Considering level structure

I’m mapping out the tree's structure, noting that in a complete binary tree, the 10th level hosts 10 leaf nodes, which could be the last level or the second to last.

Determining total nodes

I’m calculating the maximum number of nodes in a complete binary tree, considering the 10th level with 10 leaves, which might be the last level or second to last.

Assessing node maximization

OK, I’m considering how to maximize nodes by fully filling levels 1 to 9 and possibly partially filling beyond the 10th level, which likely is the last.

Mapping out tree nodes

OK, let's think this through: the 10th level with 10 leaf nodes likely denotes the last level. Having nodes beyond it would negate their leaf status.

Determining node count

Interestingly enough, a complete binary tree with 10 levels, fully filled to level 9, can have up to 1023 nodes. The 10th level adds up to 2^9 = 512 nodes.

Figuring out node capacity

For a complete binary tree with 10 levels, a fully filled 10th level would have 512 nodes. With 10 nodes on level 10, 2^9 = 512 nodes can be added.

Confirming tree structure

Interestingly enough, in a complete binary tree, level 10 must be fully filled with 512 nodes to add nodes to level 11. Therefore, level 10 contains exactly 10 nodes.

Maximizing level 9 nodes

Interestingly, filling level 9 completely allows level 10 to maintain 10 nodes, as the completeness up to level 9 doesn't interfere with the last level's node count.

Balancing node distribution

It's interesting to see that filling level 9 with 256 nodes ensures level 10 has just 10 nodes, since not all nodes in level 9 will have children.

Maximizing tree nodes

By filling level 9 first, the last level can hold exactly 10 nodes. This ensures the tree achieves its maximum potential node count.

Maximizing level 9

I’m thinking level 9 can hold 256 nodes if fully filled, allowing the last level to fit exactly 10 nodes.

I’m piecing it together

To ensure the last level has exactly 10 nodes, I’m figuring out how many nodes in level 9 need to be parents, filling left to right on level 10.

Organizing children

I’m placing 10 children in level 9, sequentially adding left and right children from available parents until all slots are filled, ensuring a complete binary tree structure.

Distributing responsibilities

I’m thinking about evenly distributing the 10 children among 5 parents, with each parent taking on 1 to 2 children. This ensures a balanced workload.

Mapping the skill

I’m figuring out that to have 10 children in level 10, 5 parents are needed to cover all the spots. This approach effectively manages the level 10 node distribution.

Clarifying node specifics

OK, let me see. The first 5 nodes in level 9 have 2 children each in level 10, totaling 10 children. The remaining 251 nodes in level 9 are leaves.

Piecing it together

I'm working through the 10th level having 10 leaves, while the 9th and other levels might also contain leaves, as complete binary trees typically have leaves at the last or second last levels.

Wondering about node count

Considering if the 10th level houses all leaves. If nodes in level 9 lack children, they are also leaves, potentially totaling 261 leaves.

Deciphering leaf ambiguity

Hmm, I’m thinking about the possibility of leaves in both levels 9 and 10 of a complete binary tree. The question might be deliberately vague or tricky.

Mapping out the solution

OK, let’s interpret the 10th level as the final one with 10 leaves. Filling levels 1 to 8 ensures the maximum number of nodes in the tree.

Maximizing tree nodes

Considering filling level 9 with as many nodes as possible from left to right, while ensuring the 10th level has exactly 10 nodes.

Pooling resources

Filling levels 1 to 8 completely uses 255 nodes. Then, by adding 255 nodes to level 9 from left to right, I dump all leftover nodes into level 9.

Filling node spaces

I’m placing 10 children in level 9, ensuring no more than 10 leaves in level 10. This results in a total of 521 nodes.

Maximizing node distribution

I realized that the optimal layout involves filling levels 1 to 8 completely and strategically placing nodes in level 9 to ensure exactly 10 leaves in level 10.

Maximizing node placement

I’m thinking of filling levels 1 to 8 completely, then strategically placing 256 nodes in level 9, ensuring exactly 10 leaves in level 10, totaling 521 nodes.

Aligning nodes

I'm thinking about filling levels 1 to 8 completely, then using 256 nodes in level 9, ensuring exactly 10 leaves in level 10, resulting in 521 nodes in total.

Validating the structure

I’m confirming that fully filling level 9 with 256 nodes and ensuring exactly 10 leaves in level 10 results in a total of 521 valid nodes, completing a complete binary tree.

Thought about 完全二叉树节点数量，持续 1m 22s

Answer

要回答这个问题，关键在于理解「完全二叉树」的定义以及如何在最末层（第10 层）只有 10 个叶子的条件下，使整棵树的节点数尽量多。

1. 完全二叉树的定义

一棵完全二叉树要求：

除最后一层外，每一层都必须是满的（即该层节点数达到最大值）。
最后一层（如果不满）里的节点必须从左到右连续排列，中间不能“空”位置。

换言之，在从上到下、从左到右的顺序中，不能跳过任何位置。

2. 题目条件的含义

题目给出：“它的第 10 层有 10 个叶子节点”。

通常我们可以理解为：

这棵树总高度是 10 层。
第 10 层上恰好有 10 个节点，并且它们全部是叶子。
并不排除第 9 层也有一些节点是叶子（因为在完全二叉树中，如果第 9 层的某些节点没有孩子，那么它们也是叶子）。
题目只明确告诉我们：第 10 层确实存在 10 个叶子。

要想让整棵树“节点总数”尽量多，就要尽可能地填满前几层。

3. 分层分析

为了便于说明，这里假设根节点为第 1 层，第 2 层是根节点的孩子，以此类推，直到第 10 层。

第 1 层到第 8 层
在完全二叉树想“最大化节点数”时，前面各层一定是满的。
- 第 1 层有 $2^{1-1} = 1$ 个节点
- 第 2 层有 $2^{2-1} = 2$ 个节点
- 第 3 层有 $2^{3-1} = 4$ 个节点
- …
- 第 8 层有 $2^{8-1} = 128$ 个节点
前 8 层总节点数为
$1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 \;=\; 2^8 - 1 \;=\; 255.$
第 9 层
对于一棵“尽量大”的完全二叉树，能填满的层就尽量填满。
- 若第 8 层有 128 个节点，则最多可以生出 $128\times 2 = 256$ 个孩子，
- 因而第 9 层“能够”拥有最多 $2^{9-1} = 256$ 个节点（即全部填满）。
填满第 9 层后，前 9 层节点总数
$255 + 256 = 511.$
第 10 层
现在已经知道第 9 层最多可以有 256 个节点。题目要求第 10 层总共有 10 个叶子节点（也即 10 个节点）。在完全二叉树中，最末层的节点分配是从左到右依次“给”父节点的左孩子、右孩子。要正好产生 10 个节点，需要：
- 让第 9 层最左边的一部分节点“各自生出孩子”，直到总计 10 个为止；
- 剩余的第 9 层节点则不再生孩子，直接成为叶子。
如果父节点都能同时拥有左右孩子，那么：
- 前 5 个父节点各有 2 个孩子，总共 $5 \times 2 = 10$ 个孩子，正好就是第 10 层的 10 个节点。
因此第 10 层共有 10 个节点。

4. 计算总节点数

前 9 层总节点数： $511$
第 10 层节点数： $10$

两者相加，即可得到整棵完全二叉树的最大节点数：

511 + 10 = 521.

因此，这棵完全二叉树在“第 10 层有 10 个叶子”的情况下，最多可以拥有 $521$ 个节点。

答案

\boxed{521}