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Code Issues Packages Projects Releases Wiki Activity

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

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nuomi1
2023-01-27 01:52:51 +08:00
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parent f73231568c
commit d76e6582fa
3 changed files with 298 additions and 3 deletions
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2
codes/swift/Package.swift
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@ -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"]),
]
)

190
codes/swift/chapter_tree/binary_search_tree.swift Normal file
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@ -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())
}
}

109
docs/chapter_tree/binary_search_tree.md
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@ -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