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9ab4b0b15c
* refactor: encode & decode Tree * style: build warning * feat: add Swift codes for array_representation_of_tree article
141 lines
3.5 KiB
Swift
141 lines
3.5 KiB
Swift
/**
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* File: array_binary_tree.swift
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* Created Time: 2023-07-23
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* Author: nuomi1 (nuomi1@qq.com)
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*/
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import utils
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/* 数组表示下的二叉树类 */
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class ArrayBinaryTree {
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private var tree: [Int?]
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/* 构造方法 */
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init(arr: [Int?]) {
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tree = arr
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}
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/* 节点数量 */
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func size() -> Int {
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tree.count
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}
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/* 获取索引为 i 节点的值 */
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func val(i: Int) -> Int? {
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// 若索引越界,则返回 null ,代表空位
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if i < 0 || i >= size() {
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return nil
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}
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return tree[i]
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}
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/* 获取索引为 i 节点的左子节点的索引 */
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func left(i: Int) -> Int {
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2 * i + 1
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}
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/* 获取索引为 i 节点的右子节点的索引 */
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func right(i: Int) -> Int {
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2 * i + 2
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}
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/* 获取索引为 i 节点的父节点的索引 */
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func parent(i: Int) -> Int {
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(i - 1) / 2
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}
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/* 层序遍历 */
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func levelOrder() -> [Int] {
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var res: [Int] = []
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// 直接遍历数组
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for i in stride(from: 0, to: size(), by: 1) {
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if let val = val(i: i) {
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res.append(val)
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}
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}
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return res
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}
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/* 深度优先遍历 */
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private func dfs(i: Int, order: String, res: inout [Int]) {
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// 若为空位,则返回
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guard let val = val(i: i) else {
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return
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}
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// 前序遍历
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if order == "pre" {
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res.append(val)
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}
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dfs(i: left(i: i), order: order, res: &res)
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// 中序遍历
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if order == "in" {
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res.append(val)
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}
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dfs(i: right(i: i), order: order, res: &res)
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// 后序遍历
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if order == "post" {
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res.append(val)
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}
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}
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/* 前序遍历 */
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func preOrder() -> [Int] {
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var res: [Int] = []
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dfs(i: 0, order: "pre", res: &res)
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return res
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}
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/* 中序遍历 */
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func inOrder() -> [Int] {
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var res: [Int] = []
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dfs(i: 0, order: "in", res: &res)
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return res
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}
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/* 后序遍历 */
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func postOrder() -> [Int] {
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var res: [Int] = []
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dfs(i: 0, order: "post", res: &res)
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return res
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}
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}
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@main
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enum _ArrayBinaryTree {
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/* Driver Code */
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static func main() {
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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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let arr = [1, 2, 3, 4, nil, 6, 7, 8, 9, nil, nil, 12, nil, nil, 15]
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let root = TreeNode.listToTree(arr: arr)
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print("\n初始化二叉树\n")
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print("二叉树的数组表示:")
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print(arr)
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print("二叉树的链表表示:")
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PrintUtil.printTree(root: root)
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// 数组表示下的二叉树类
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let abt = ArrayBinaryTree(arr: arr)
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// 访问节点
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let i = 1
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let l = abt.left(i: i)
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let r = abt.right(i: i)
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let p = abt.parent(i: i)
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print("\n当前节点的索引为 \(i) ,值为 \(abt.val(i: i) as Any)")
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print("其左子节点的索引为 \(l) ,值为 \(abt.val(i: l) as Any)")
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print("其右子节点的索引为 \(r) ,值为 \(abt.val(i: r) as Any)")
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print("其父节点的索引为 \(p) ,值为 \(abt.val(i: p) as Any)")
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// 遍历树
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var res = abt.levelOrder()
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print("\n层序遍历为:\(res)")
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res = abt.preOrder()
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print("前序遍历为:\(res)")
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res = abt.inOrder()
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print("中序遍历为:\(res)")
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res = abt.postOrder()
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print("后序遍历为:\(res)")
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}
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}
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