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完全二叉树节点数量

생성일: 2025년 1월 20일

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질문

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？

사고

Mapping the tree dynamics

I'm piecing together a complete binary tree with 10th level having 10 leaf nodes, analyzing node distribution through adjacency and subtree structures.

Calculating 9th level nodes

OK, let me see. The 9th level has 256 nodes, ensuring the 10th level accommodates 10 leaf nodes within a complete binary tree structure.

Charting node distribution

I’m thinking through how the 9th level in a complete binary tree with 10 levels is fully populated, leading to 511 nodes before adding the 10th level's 10 nodes.

Pulling together

I’m summing 511 nodes from the first 9 levels with 10 nodes on the 10th level, potentially totaling 521, while considering the possibility of missing nodes from the 9th level.

Balancing node distribution

Maximizing tree nodes by ensuring 10 leaves on the 10th level, while only having 10 parent nodes in the 9th level. Each parent can have up to 2 children.

Maximizing node count

I’m figuring out how to fit 10 leaves on level 10 while filling levels 1-8. The challenge is arranging nodes in level 9 to ensure we meet this node configuration.

Balancing node placement

I'm trying to max out nodes in level 9 while ensuring the 10th level has exactly 10 children, using a left-to-right filling method for the complete binary tree.

Ensuring proper configurations

In a complete binary tree, nodes typically have both children. For 10 nodes in the last level, they must be distributed left to right among the level 9 parents.

Laying out the options

I'm thinking through whether distributing 10 children among level 9 parents to maximize nodes is the best approach. This involves evaluating the total number of nodes in the tree.

Strategizing node allocation

OK, let me see. I’m figuring out how to place 10 children in level 10 while placing as many nodes as possible in level 9, ensuring the last level is filled from left to right.

Assigning children to parents

I'm piecing together how to distribute 10 children among parents in level 9, filling the last level from left to right, starting with the leftmost parent.

Assigning remaining children

OK, let me see: I’m assigning the last two children to the fifth parent in level 9, ensuring all 10 children are allocated while filling the last level from left to right.

Balancing node allocation

OK, let me see. Level 9 parents without children don't contradict the completeness of a binary tree, as the requirement only applies to levels 1 through 10, with the last level filled left to right.

Evaluating node assignments

Presence without children in level 9 holds no issue with tree completeness, given the flexibility for nodes with right children to lack left children.

Mapping the constraints

In a complete binary tree, skipping a left child in level 9 prevents any parent node to the right from having children, thus maintaining full left-to-right coverage.

Identifying remaining children

I’m tracing the 8 remaining children in level 9, ensuring the last level fills from left to right up to exactly 10 children.

Maximizing node allocation

I’m assigning 10 children to 5 parents in level 9, ensuring the 10th level fills from left to right, and confirming the maximal number of nodes without gaps.

Mapping node assignments

OK, let me see. The 5 parents in level 9 each get 2 children, making 10 children in total. No gaps exist in the last level.

Confirming a valid tree

I’m concluding that the complete binary tree is valid with the first 5 parents in level 9 having 2 children each, leading to a total of 10 children filling the first 10 positions in the last level.

Confirming tree layout

I’m piecing together that arranging the first 5 parents in level 9 with 2 children each fills the first 10 positions in the last level, totaling 521 nodes.

Confirming completeness

I’m verifying that levels 1 to 9 are fully populated, with 511 nodes, while the 10th level has at least one node. This ensures the tree meets completeness criteria.

Assessing node distribution

Checking the first 9 levels host 511 nodes fully, with 256 nodes in level 9 alone. The 10th level has 10 leaves.

Thought about 完全二叉树节点数，持续 2m 20s

답변

分析如下：

  1. 完全二叉树的定义

    • 除了最末层之外，每一层都是“满的”；
    • 最末层的结点尽量靠左对齐。

  2. 已知条件

    • 这棵完全二叉树共有 10 层；
    • 第 10 层（最末层）只有 10 个叶子结点。

  3. 思路概述

    • 若一棵完全二叉树有 hh 层，则第 h−1h-1 层之前（即 1～h−1h-1 层）都必须是满的；
    • 在第 hh 层中，结点自左向右依次存在，中间不能“空”出位置。

    由于本题的高度是 10，那么 1～9 层都必须是满的。令第 9 层的结点数为 N9N_9，它将全部存在（因为第 9 层不是最末层），而第 10 层部分填满，自左至右仅有 10 个结点。

  4. 计算前 9 层的结点数

    • 第 kk 层有 2k−12^{k-1} 个结点；
    • 前 9 层的总结点数为 1+2+4+⋯+28  =  29−1  =  511. 1 + 2 + 4 + \dots + 2^{8} \;=\; 2^{9} - 1 \;=\; 511.

  5. 第 10 层如何放 10 个结点才能保证整棵树“最大”？

    • 第 9 层有 28=2562^8 = 256 个结点，都是“父亲”候选；
    • 在完全二叉树中，最末层从左至右依次放结点：
      1. 第 9 层的第 1 个结点，若还有可用的名额（叶子要放），它会先放左孩子，再放右孩子；
      2. 第 9 层的第 2 个结点也是如此……
    • 由于总共只能放 10 个叶子，那么前面若有 5 个父结点各放 2 个孩子，就正好得到 10 个孩子；之后的父结点即使存在，也“没有多余名额”再放孩子了。
    • 这样既满足了“自左向右”放置叶子的要求，又不会浪费任何可能的父结点位置。

  6. 最终总结点数

    • 前 9 层共有 511 个结点；
    • 第 10 层有 10 个结点；
    • 因此这棵完全二叉树最多可拥有 511+10=521 511 + 10 = 521 个结点。

答案：521

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