完全二叉树最大节点数

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

Identifying key elements

I'm thinking through the data points. It’s a full binary tree at the 10th level with exactly 10 leaf nodes.

Calculating nodes

A complete binary tree with 10 levels typically has 1023 nodes. With only 10 leaf nodes at the 10th level, the first 9 levels are fully populated, totaling 511 nodes.

Redistributing nodes

OK, I’m thinking about the 10th level node count. With 10 leaf nodes, the remaining 511 nodes from the first 9 levels need reallocation to use 521 nodes total.

Maximizing nodes

Partitioning the 521 nodes, 511 go to the first 9 levels and 10 to the 10th level. This ensures full coverage of all levels.

Confirming node placement

I’m ensuring 511 nodes fill the first 9 levels, with 10 nodes in the 10th level, all as leaves. This configuration meets the condition of having 10 leaf nodes.

Ensuring optimal placement

I’m focusing on maximizing nodes in levels 1 to 9 by keeping them fully populated, with 10 nodes in the 10th level as leaves, totaling 521 nodes.

Confirming placement

I’m piecing together that the 10th level has 10 nodes, with the leftmost 5 nodes in the 9th level having 2 children each, matching the complete binary tree definition.

Thought about 完全二叉树节点，持续 32 秒

回答

题意可以理解为：这是一棵“完全二叉树”，它一共有 10 层，并且第 10 层恰好有 10 个结点（全部是叶子结点）。问在此条件下，这棵树可能拥有的最大结点数是多少？

分析思路

完全二叉树的定义：
- 从第 1 层（根节点所在层）到倒数第 2 层都必须是“满的”，也就是对应层上的所有节点全部存在。
- 最后一层（这里是第 10 层）的节点也尽量“左对齐”地排列，不能在中间或右侧空出位置再出现节点。
若前 9 层都“满”，则前 9 层节点总数为：
$2^0 + 2^1 + 2^2 + \dots + 2^8 = 2^9 - 1 = 511$
第 10 层“只有”10 个节点，说明第 10 层并未满（因为若满应有 $2^9 = 512$ 个节点）。但为了让整棵树的节点总数尽量大，前 9 层依然应尽可能地满，即保留这 511 个节点，然后在第 10 层从左到右放入 10 个节点。

因此整棵树的最大节点数即为

\underbrace{511}_{\text{前 9 层}} + \underbrace{10}_{\text{第 10 层}} \;=\; 521.

结论

这棵完全二叉树最多可以拥有 521 个节点。

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