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* Fix is_empty() implementation in the stack and queue chapter * Update en/CONTRIBUTING.md * Remove "剩余" from the state definition of knapsack problem * Sync zh and zh-hant versions * Update the stylesheets of code tabs * Fix quick_sort.rb * Fix TS code * Update chapter_paperbook * Upload the manuscript of 0.1 section * Fix binary_tree_dfs.rb * Bug fixes * Update README * Update README * Update README * Update README.md * Update README * Sync zh and zh-hant versions * Bug fixes
161 lines
3.6 KiB
Ruby
161 lines
3.6 KiB
Ruby
=begin
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File: binary_search_tree.rb
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Created Time: 2024-04-18
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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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### 二元搜尋樹 ###
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class BinarySearchTree
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### 建構子 ###
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def initialize
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# 初始化空樹
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@root = nil
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end
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### 獲取二元樹根節點 ###
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def get_root
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@root
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end
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### 查詢節點 ###
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def search(num)
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cur = @root
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# 迴圈查詢,越過葉節點後跳出
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while !cur.nil?
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# 目標節點在 cur 的右子樹中
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if cur.val < num
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cur = cur.right
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# 目標節點在 cur 的左子樹中
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elsif cur.val > num
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cur = cur.left
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# 找到目標節點,跳出迴圈
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else
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break
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end
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end
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cur
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end
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### 插入節點 ###
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def insert(num)
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# 若樹為空,則初始化根節點
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if @root.nil?
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@root = TreeNode.new(num)
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return
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end
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# 迴圈查詢,越過葉節點後跳出
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cur, pre = @root, nil
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while !cur.nil?
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# 找到重複節點,直接返回
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return if cur.val == num
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pre = cur
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# 插入位置在 cur 的右子樹中
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if cur.val < num
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cur = cur.right
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# 插入位置在 cur 的左子樹中
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else
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cur = cur.left
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end
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end
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# 插入節點
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node = TreeNode.new(num)
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if pre.val < num
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pre.right = node
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else
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pre.left = node
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end
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end
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### 刪除節點 ###
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def remove(num)
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# 若樹為空,直接提前返回
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return if @root.nil?
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# 迴圈查詢,越過葉節點後跳出
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cur, pre = @root, nil
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while !cur.nil?
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# 找到待刪除節點,跳出迴圈
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break if cur.val == num
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pre = cur
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# 待刪除節點在 cur 的右子樹中
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if cur.val < num
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cur = cur.right
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# 待刪除節點在 cur 的左子樹中
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else
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cur = cur.left
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end
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end
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# 若無待刪除節點,則直接返回
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return if cur.nil?
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# 子節點數量 = 0 or 1
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if cur.left.nil? || cur.right.nil?
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# 當子節點數量 = 0 / 1 時, child = null / 該子節點
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child = cur.left || cur.right
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# 刪除節點 cur
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if cur != @root
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if pre.left == cur
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pre.left = child
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else
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pre.right = child
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end
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else
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# 若刪除節點為根節點,則重新指定根節點
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@root = child
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end
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# 子節點數量 = 2
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else
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# 獲取中序走訪中 cur 的下一個節點
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tmp = cur.right
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while !tmp.left.nil?
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tmp = tmp.left
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end
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# 遞迴刪除節點 tmp
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remove(tmp.val)
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# 用 tmp 覆蓋 cur
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cur.val = tmp.val
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end
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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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# 初始化二元搜尋樹
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bst = BinarySearchTree.new
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nums = [8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15]
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# 請注意,不同的插入順序會生成不同的二元樹,該序列可以生成一個完美二元樹
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nums.each { |num| bst.insert(num) }
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puts "\n初始化的二元樹為\n"
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print_tree(bst.get_root)
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# 查詢節點
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node = bst.search(7)
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puts "\n查詢到的節點物件為: #{node},節點值 = #{node.val}"
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# 插入節點
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bst.insert(16)
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puts "\n插入節點 16 後,二元樹為\n"
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print_tree(bst.get_root)
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# 刪除節點
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bst.remove(1)
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puts "\n刪除節點 1 後,二元樹為\n"
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print_tree(bst.get_root)
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bst.remove(2)
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puts "\n刪除節點 2 後,二元樹為\n"
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print_tree(bst.get_root)
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bst.remove(4)
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puts "\n刪除節點 4 後,二元樹為\n"
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print_tree(bst.get_root)
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end
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