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匿名使用者 發問時間: 電腦與網際網路程式設計 · 1 0 年前

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Define the Fibonanci binary tree of order n as follows:

█ If n=0 or 1, the tree consists of a single node.

█ If n>1, the tree consists of a root, with the Fibonanci tree of order n-1 as the left subtree and the Fibonanci tree of order n-2 as the right subtree.

1. What's the number of leaves in the Fibonanci tree of order n? Why?

2. What's the depth of Fibonanci tree of order n?

2 個解答

評分
  • 鴨子
    Lv 6
    1 0 年前
    最佳解答

    1.

    the number of leaves in the Fibonanci tree of order n

    = Fibonanci 數列的 第 n+1 項

    採標準Fibonanci 數列的定義

    F(0)=0

    F(1)=1

    F(n)=F(n-1)+F(n-2), n>1

    那答案為Fibonanci 數列的第 n+1 項

    ==========================

    理由:

    Leaf 就是沒有左樹,右樹的 node

    a binary tree (如果不是leaf) 的 leaf數

    =左樹的 leaf數 + 右樹的 leaf數

    (因為根不是leaf

    a binary tree (是leaf)的 leaf數

    =1

    定義: L(n) = the number of leaves in the Fibonanci tree of order n

    L(0)=1 (因為 If n=0 or 1, the tree consists of a single node.)

    L(1)=1

    L(n)=L(n-1)+L(n-2)

    因為If n>1, the tree consists of a root, with the Fibonanci tree of order n-1 as the left subtree and the Fibonanci tree of order n-2 as the right subtree.

    =>

    L(n) 是 Fibonanci 數列的第n項

    =========================================================

    What's the depth of Fibonanci tree of order n?

    n=0, 為 0

    n>0, 為 n-1

    因為:

    The depth of a node n is the length of the path from the root to the node.

    The root node is at depth zero.

    D(0)=0

    D(1)=0

    D(2)=1

    D(3)=2

    D(4)=3

    D(5)=4

    .....

    D(n)=0, n= 0

    D(n)= n-1, n>0

  • 6 年前

    硬碟要找硬碟醫院,能救回來才重要,我司就是找他們做出來的

    必須要推,當初心急亂找錯店告訴我沒辦法,後來找又找到硬碟醫院,我告訴他真心要來救援,可以等,在他們努力下有找回來

    太棒了,原來微軟張經理都找過他們

    http://www.datamaster.com.tw/

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