完全二叉树节点数

Creado el: 20 de enero de 2025

Respondido usando O1 por Chat01

Pregunta

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?

Pensamiento

Understanding the tree structure

OK, let me see. The problem involves a fully filled binary tree with a 10th level having 10 leaf nodes. The task is to determine the maximum number of nodes in the tree, considering the constraints.

Understanding the structure

A complete binary tree of height H usually spans levels 1 to H. Since the 10th level has only 10 nodes, it is partially filled, affecting the total node count.

Maximizing node distribution

I’m pulling together insights on distributing nodes efficiently. By optimizing the 9th level to hold 256 nodes and filling 10 nodes in the 10th level, the total number of nodes is maximized.

Confirming tree structure

I'm verifying that the 9th level is completely filled with 256 nodes, and the 10th level has 10 nodes, partially completing the complete binary tree.

Calculating the total

Adding the nodes from levels 1 to 9 gives 511, plus 10 nodes from level 10, totaling 521 nodes. This confirms the maximum node count.

Verifying results

I'm confirming the total count of 521 nodes by using the complete binary tree formula, ensuring precision in the calculation for the given height and nodes.

Thought about 完全二叉树节点数,持续 31 秒

Respuesta

要得到最大节点数,关键在于「完全二叉树」的定义:除最后一层外,每一层都必须被填满,且最后一层的节点从左向右连续排列。题目中第 10 层只有 10 个叶子节点,因此前 9 层必然是满的。

  • 第 1 层(根节点)有 1=201 = 2^0 个节点
  • 第 2 层有 2=212 = 2^1 个节点
  • 第 3 层有 4=224 = 2^2 个节点
  • \dots
  • 第 9 层有 256=28256 = 2^8 个节点

前 9 层总节点数为

1+2+4++256=291=511.1 + 2 + 4 + \cdots + 256 = 2^9 - 1 = 511.

由于第 10 层只剩 10 个叶子节点,所以整棵树的最大节点数为

511+10=521.511 + 10 = 521.

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