feat: add Swift codes for binary_search_tree article (#296)

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"]),
     ]
 )

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())
+    }
+}

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