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125 lines
2.5 KiB
Ruby
125 lines
2.5 KiB
Ruby
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=begin
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File: array_binary_tree.rb
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Created Time: 2024-04-17
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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require_relative '../utils/tree_node'
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require_relative '../utils/print_util'
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### 数组表示下的二叉树类 ###
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class ArrayBinaryTree
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### 构造方法 ###
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def initialize(arr)
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@tree = arr.to_a
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end
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### 列表容量 ###
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def size
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@tree.length
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end
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### 获取索引为 i 节点的值 ###
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def val(i)
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# 若索引越界,则返回 nil ,代表空位
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return if i < 0 || i >= size
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@tree[i]
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end
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### 获取索引为 i 节点的左子节点的索引 ###
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def left(i)
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2 * i + 1
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end
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### 获取索引为 i 节点的右子节点的索引 ###
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def right(i)
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2 * i + 2
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end
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### 获取索引为 i 节点的父节点的索引 ###
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def parent(i)
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(i - 1) / 2
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end
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### 层序遍历 ###
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def level_order
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@res = []
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# 直接遍历数组
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for i in 0...size
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@res << val(i) unless val(i).nil?
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end
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@res
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end
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### 深度优先遍历 ###
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def dfs(i, order)
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return if val(i).nil?
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# 前序遍历
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@res << val(i) if order == :pre
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dfs(left(i), order)
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# 中序遍历
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@res << val(i) if order == :in
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dfs(right(i), order)
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# 后序遍历
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@res << val(i) if order == :post
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end
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### 前序遍历 ###
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def pre_order
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@res = []
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dfs(0, :pre)
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@res
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end
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### 中序遍历 ###
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def in_order
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@res = []
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dfs(0, :in)
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@res
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end
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### 后序遍历 ###
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def post_order
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@res = []
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dfs(0, :post)
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@res
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end
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end
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### Driver Code ###
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if __FILE__ == $0
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# 初始化二叉树
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# 这里借助了一个从数组直接生成二叉树的函数
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arr = [1, 2, 3, 4, nil, 6, 7, 8, 9, nil, nil, 12, nil, nil, 15]
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root = arr_to_tree(arr)
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puts "\n初始化二叉树\n\n"
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puts '二叉树的数组表示:'
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pp arr
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puts '二叉树的链表表示:'
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print_tree(root)
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# 数组表示下的二叉树类
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abt = ArrayBinaryTree.new(arr)
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# 访问节点
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i = 1
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l, r, _p = abt.left(i), abt.right(i), abt.parent(i)
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puts "\n当前节点的索引为 #{i} ,值为 #{abt.val(i).inspect}"
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puts "其左子节点的索引为 #{l} ,值为 #{abt.val(l).inspect}"
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puts "其右子节点的索引为 #{r} ,值为 #{abt.val(r).inspect}"
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puts "其父节点的索引为 #{_p} ,值为 #{abt.val(_p).inspect}"
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# 遍历树
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res = abt.level_order
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puts "\n层序遍历为: #{res}"
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res = abt.pre_order
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puts "前序遍历为: #{res}"
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res = abt.in_order
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puts "中序遍历为: #{res}"
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res = abt.post_order
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puts "后序遍历为: #{res}"
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end
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