From d76e6582fa04e595930b26c412adba721605de0a Mon Sep 17 00:00:00 2001 From: nuomi1 Date: Fri, 27 Jan 2023 01:52:51 +0800 Subject: [PATCH] feat: add Swift codes for binary_search_tree article (#296) --- codes/swift/Package.swift | 2 + .../chapter_tree/binary_search_tree.swift | 190 ++++++++++++++++++ docs/chapter_tree/binary_search_tree.md | 109 +++++++++- 3 files changed, 298 insertions(+), 3 deletions(-) create mode 100644 codes/swift/chapter_tree/binary_search_tree.swift diff --git a/codes/swift/Package.swift b/codes/swift/Package.swift index 8622baf3..1dc64032 100644 --- a/codes/swift/Package.swift +++ b/codes/swift/Package.swift @@ -25,6 +25,7 @@ let package = Package( .executable(name: "binary_tree", targets: ["binary_tree"]), .executable(name: "binary_tree_bfs", targets: ["binary_tree_bfs"]), .executable(name: "binary_tree_dfs", targets: ["binary_tree_dfs"]), + .executable(name: "binary_search_tree", targets: ["binary_search_tree"]), ], targets: [ .target(name: "utils", path: "utils"), @@ -48,5 +49,6 @@ let package = Package( .executableTarget(name: "binary_tree", dependencies: ["utils"], path: "chapter_tree", sources: ["binary_tree.swift"]), .executableTarget(name: "binary_tree_bfs", dependencies: ["utils"], path: "chapter_tree", sources: ["binary_tree_bfs.swift"]), .executableTarget(name: "binary_tree_dfs", dependencies: ["utils"], path: "chapter_tree", sources: ["binary_tree_dfs.swift"]), + .executableTarget(name: "binary_search_tree", dependencies: ["utils"], path: "chapter_tree", sources: ["binary_search_tree.swift"]), ] ) diff --git a/codes/swift/chapter_tree/binary_search_tree.swift b/codes/swift/chapter_tree/binary_search_tree.swift new file mode 100644 index 00000000..9d5025c0 --- /dev/null +++ b/codes/swift/chapter_tree/binary_search_tree.swift @@ -0,0 +1,190 @@ +/** + * File: binary_search_tree.swift + * Created Time: 2023-01-26 + * Author: nuomi1 (nuomi1@qq.com) + */ + +import utils + +/* 二叉搜索树 */ +class BinarySearchTree { + private var root: TreeNode? + + init(nums: [Int]) { + let nums = nums.sorted() // 排序数组 + root = buildTree(nums: nums, i: 0, j: nums.count - 1) // 构建二叉搜索树 + } + + /* 获取二叉树根结点 */ + func getRoot() -> TreeNode? { + root + } + + /* 构建二叉搜索树 */ + func buildTree(nums: [Int], i: Int, j: Int) -> TreeNode? { + if i > j { + return nil + } + // 将数组中间结点作为根结点 + let mid = (i + j) / 2 + let root = TreeNode(x: nums[mid]) + // 递归建立左子树和右子树 + root.left = buildTree(nums: nums, i: i, j: mid - 1) + root.right = buildTree(nums: nums, i: mid + 1, j: j) + return root + } + + /* 查找结点 */ + func search(num: Int) -> TreeNode? { + var cur = root + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 目标结点在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 目标结点在 cur 的左子树中 + else if cur!.val > num { + cur = cur?.left + } + // 找到目标结点，跳出循环 + else { + break + } + } + // 返回目标结点 + return cur + } + + /* 插入结点 */ + func insert(num: Int) -> TreeNode? { + // 若树为空，直接提前返回 + if root == nil { + return nil + } + var cur = root + var pre: TreeNode? + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 找到重复结点，直接返回 + if cur!.val == num { + return nil + } + pre = cur + // 插入位置在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 插入位置在 cur 的左子树中 + else { + cur = cur?.left + } + } + // 插入结点 val + let node = TreeNode(x: num) + if pre!.val < num { + pre?.right = node + } else { + pre?.left = node + } + return node + } + + /* 删除结点 */ + @discardableResult + func remove(num: Int) -> TreeNode? { + // 若树为空，直接提前返回 + if root == nil { + return nil + } + var cur = root + var pre: TreeNode? + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 找到待删除结点，跳出循环 + if cur!.val == num { + break + } + pre = cur + // 待删除结点在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 待删除结点在 cur 的左子树中 + else { + cur = cur?.left + } + } + // 若无待删除结点，则直接返回 + if cur == nil { + return nil + } + // 子结点数量 = 0 or 1 + if cur?.left == nil || cur?.right == nil { + // 当子结点数量 = 0 / 1 时， child = null / 该子结点 + let child = cur?.left != nil ? cur?.left : cur?.right + // 删除结点 cur + if pre?.left === cur { + pre?.left = child + } else { + pre?.right = child + } + } + // 子结点数量 = 2 + else { + // 获取中序遍历中 cur 的下一个结点 + let nex = getInOrderNext(root: cur?.right) + let tmp = nex!.val + // 递归删除结点 nex + remove(num: nex!.val) + // 将 nex 的值复制给 cur + cur?.val = tmp + } + return cur + } + + /* 获取中序遍历中的下一个结点（仅适用于 root 有左子结点的情况） */ + func getInOrderNext(root: TreeNode?) -> TreeNode? { + var root = root + if root == nil { + return root + } + // 循环访问左子结点，直到叶结点时为最小结点，跳出 + while root?.left != nil { + root = root?.left + } + return root + } +} + +@main +enum _BinarySearchTree { + /* Driver Code */ + static func main() { + /* 初始化二叉搜索树 */ + let nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15] + let bst = BinarySearchTree(nums: nums) + print("\n初始化的二叉树为\n") + PrintUtil.printTree(root: bst.getRoot()) + + /* 查找结点 */ + var node = bst.search(num: 7) + print("\n查找到的结点对象为 \(node!)，结点值 = \(node!.val)") + + /* 插入结点 */ + node = bst.insert(num: 16) + print("\n插入结点 16 后，二叉树为\n") + PrintUtil.printTree(root: bst.getRoot()) + + /* 删除结点 */ + bst.remove(num: 1) + print("\n删除结点 1 后，二叉树为\n") + PrintUtil.printTree(root: bst.getRoot()) + bst.remove(num: 2) + print("\n删除结点 2 后，二叉树为\n") + PrintUtil.printTree(root: bst.getRoot()) + bst.remove(num: 4) + print("\n删除结点 4 后，二叉树为\n") + PrintUtil.printTree(root: bst.getRoot()) + } +} diff --git a/docs/chapter_tree/binary_search_tree.md b/docs/chapter_tree/binary_search_tree.md index 75ffb678..09324ae4 100644 --- a/docs/chapter_tree/binary_search_tree.md +++ b/docs/chapter_tree/binary_search_tree.md @@ -191,7 +191,27 @@ comments: true === "Swift" ```swift title="binary_search_tree.swift" - + /* 查找结点 */ + func search(num: Int) -> TreeNode? { + var cur = root + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 目标结点在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 目标结点在 cur 的左子树中 + else if cur!.val > num { + cur = cur?.left + } + // 找到目标结点，跳出循环 + else { + break + } + } + // 返回目标结点 + return cur + } ``` ### 插入结点 @@ -428,7 +448,39 @@ comments: true === "Swift" ```swift title="binary_search_tree.swift" - + /* 插入结点 */ + func insert(num: Int) -> TreeNode? { + // 若树为空，直接提前返回 + if root == nil { + return nil + } + var cur = root + var pre: TreeNode? + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 找到重复结点，直接返回 + if cur!.val == num { + return nil + } + pre = cur + // 插入位置在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 插入位置在 cur 的左子树中 + else { + cur = cur?.left + } + } + // 插入结点 val + let node = TreeNode(x: num) + if pre!.val < num { + pre?.right = node + } else { + pre?.left = node + } + return node + } ``` 为了插入结点，需要借助 **辅助结点 `prev`** 保存上一轮循环的结点，这样在遍历到 $\text{null}$ 时，我们也可以获取到其父结点，从而完成结点插入操作。 @@ -820,7 +872,58 @@ comments: true === "Swift" ```swift title="binary_search_tree.swift" - + /* 删除结点 */ + @discardableResult + func remove(num: Int) -> TreeNode? { + // 若树为空，直接提前返回 + if root == nil { + return nil + } + var cur = root + var pre: TreeNode? + // 循环查找，越过叶结点后跳出 + while cur != nil { + // 找到待删除结点，跳出循环 + if cur!.val == num { + break + } + pre = cur + // 待删除结点在 cur 的右子树中 + if cur!.val < num { + cur = cur?.right + } + // 待删除结点在 cur 的左子树中 + else { + cur = cur?.left + } + } + // 若无待删除结点，则直接返回 + if cur == nil { + return nil + } + // 子结点数量 = 0 or 1 + if cur?.left == nil || cur?.right == nil { + // 当子结点数量 = 0 / 1 时， child = null / 该子结点 + let child = cur?.left != nil ? cur?.left : cur?.right + // 删除结点 cur + if pre?.left === cur { + pre?.left = child + } else { + pre?.right = child + } + } + // 子结点数量 = 2 + else { + // 获取中序遍历中 cur 的下一个结点 + let nex = getInOrderNext(root: cur?.right) + let tmp = nex!.val + // 递归删除结点 nex + remove(num: nex!.val) + // 将 nex 的值复制给 cur + cur?.val = tmp + } + return cur + } ``` ## 二叉搜索树的优势