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完全二叉树节点数量

Created on: January 20, 2025

Answered using O1 by Chat01

Question

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？

Thought

Determining the maximum nodes

OK, let me see. Considering the tree's structure and the 10th level's 10 leaf nodes, it seems the tree could have as many as 767 nodes at most.

Maximizing the tree

I’m working out the tree's structure by filling levels 1 to 9 completely, totaling 511 nodes, then figuring out how many more nodes fit into the 10th level.

Figuring out node distribution

I'm thinking through the 10th level nodes, where 10 leaf nodes suggest at least 5 nodes, considering optimal layout and complete tree rules.

Confirming node placement

I’m verifying that the 10th level's 10 leaf nodes are situated in the leftmost positions, with all 256 nodes from level 9 fully utilized to maximize the total node count.

Assigning children

OK, let me see. I’m figuring out how to assign 10 children to the 9th level nodes, ensuring the 10th level's 10 leftmost positions are fully occupied.

Ensuring node allocation

I’m checking the 10th level has 10 nodes by giving the first 5 nodes in level 9 exactly 2 children each, filling the leftmost 10 spots in the 10th level.

Confirming node calculation

I’m verifying that the total nodes, including 10 at level 10, are 521. Reassessing reveals this number fits the complete binary tree's structure.

Thinking through tree structure

I'm exploring if the 10th level is the last or if more levels are possible. Considering node distribution and tree completeness, I aim to understand the implications for the structure.

Understanding level utilization

Hmm, I'm thinking about how in a complete binary tree, unused spots in earlier levels prevent adding nodes to further levels. This keeps the structure neat and balanced.

Piecing it together

Interestingly enough, the 10th level contains 10 leaf nodes, making it the tree's final level. The earlier 9 levels are completely filled with a total of 511 nodes.

Maximizing distribution

I'm thinking about how to evenly distribute the 10 children among 256 parents in level 9 of the complete binary tree, ensuring completeness and maximizing the tree's total nodes.

Distributing children

I’m focusing on how 10 children are evenly assigned to 5 parents on level 9, ensuring completeness and filling the leftmost 10 spots on level 10, totaling 521 nodes.

Ensuring distribution accuracy

Mapping out child allocation to parents in level 9 confirms no left-to-right imbalance, ensuring 5 parents have 2 children each and the 10th child completes level 10.

Thought about 完全二叉树节点数量，持续 1m 25s

Answer

题意是：有一棵高度为 10 的完全二叉树，它的第 10 层并未被填满，仅有 10 个叶子节点，问在这种情况下，这棵树最多可以有多少个节点。

分析思路

  1. 完全二叉树 的定义要求：从上到下、从左到右逐层“尽量”填满，上一层若未满，则下一层不能有节点；同一层中若有空位，则只能出现在该层最右侧位置。

  2. 前 9 层已满

    • 在为了“尽量多”节点的情况下，树的前 9 层都会被填满。
    • 前 9 层节点数总和为 20+21+22+⋯+28  =  29−1  =  511.2^0 + 2^1 + 2^2 + \cdots + 2^8 \;=\; 2^9 - 1 \;=\; 511.

  3. 第 10 层有 10 个叶子节点

    • 在第 10 层出现的节点，全部是叶子（因为题意说“只有 10 个叶子结点在第 10 层”，也就意味着第 10 层没有再往下的子节点）。
    • 由于是完全二叉树，第 10 层的这 10 个节点必须从左到右依次填满前 10 个位置。

  4. 如何得到第 10 层的 10 个节点？

    • 第 9 层共有 28=2562^8 = 256 个节点（因为前 8 层是满的），每个节点可以有 0、1 或 2 个孩子。
    • 为了让第 10 层恰好出现 10 个节点，前若干个第 9 层的节点合计提供了这 10 个子节点，然后其余的第 9 层节点不再有子节点。
    • 满足完全二叉树“从左到右填充”的要求时，只要前几个第 9 层节点合计生成这 10 个孩子，第 10 层就会顺次占据最左侧的 10 个位置。

  5. 总节点数计算

    • 前 9 层满：511511 节点
    • 第 10 层：1010 节点
    • 合计：511+10=521511 + 10 = 521

结论

这棵完全二叉树在第 10 层只有 10 个叶子节点的情况下，为了使整棵树的节点数尽量多，前 9 层全部填满，第 10 层填左侧 10 个节点。由此可得 最多节点数是 521。

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