前言

這題是極其麻煩極其麻煩的一道題（前提是你不知道它有套路）……
我們不講那些歪門邪道，我們正兒八經的解一下，想正經求解，很麻煩很麻煩。。。

題目要求

P1305題題解

分析

這題你看著容易，那是你想當然認為建立一棵樹，給你的序列就是先根的，其實真實情況里不一定是……（當然本題是，所以是橙題……）

這題肝的我簡直要暴斃，要多麻煩有多麻煩，大家如果有改進策略請告訴我，我會虛心接受了。。。寫這么長代碼累死人啦！！

我講講我分析的思路吧：

我們要自己編寫結點類，其實結點湊起來就是tree，我們控一下root就可以咯。
手寫Node，由于指定類型，所以不加泛型，不加setter、getter，玩最簡單的版本：

static class Node {char data;Node left, right;Node(char data) {this.data = data;this.left = null;this.right = null;}@Overridepublic boolean equals(Object o) {if (this == o) return true;if (o == null || getClass() != o.getClass()) return false;Node node = (Node) o;return data == node.data;}@Overridepublic int hashCode() {return Objects.hash(data);}}

對咯，這個必須是（靜態）內部類，因為luoguOJ不識別，你寫成另一個class會判CE的。。。。。。靜態的內部類是OK的，因為靜態不能訪問非靜態的，這會給方法里的調用帶來麻煩。
還有就是一定要重寫hashCode()和equals()，我們可是需要用data來區別Node的呀！！

首先我們建立一棵樹，需要建立臨時根結點，由于我們想知道這個結點是否已經在樹里面，我們不想再次遍歷，那我們可以建立一個HashMap，存儲已經加入樹中的結點。
我們還需要一個HashMap，用來為未插入樹中的結點提供查找的便利。

將初始化的根結點保存為根結點，左右兒子加上去。不過這個題無論何時都要注意 ‘*’ 的存在，必須不能插入data為 ‘*’ 的結點。

接下來開循環，每次讀一行數據，然后切成三份(toCharArray())，用三個char存儲。
循環里面先看看這個臨時的根結點是不是在map里，如果有就說明它在樹里，它不可能還有左右兒子，所以肯定是原先作為某個子結點的，進行一下替換。
如果不在map里，就去unAddedMap里找，這里有的話思路同上，只不過這是一群沒加入樹里的臨時樹，相當于與真正的樹構成了一個森林而已……
如果都不在，那好，這就是沒有出現過的結點，我們看看它是不是新的根結點，這個判斷要用臨時根結點的左右兒子分別與當前真實根節點比較，如果equals就換根，注意換根以后要查右兒子在不在unAddedMap里（肯定不在map里）。。如果不是換根的情況那就只能把它塞進unAddedMap里了，畢竟這在當前是一棵 “孤兒樹”，無依無靠嗚嗚嗚~~

大致是這樣，具體的看我代碼注釋吧，能看完說明你是個狠人，哦不，狼人（比狠人還狠一、）。。。

哦哦哦，差點忘說了，這個算法有一部分的邏輯是這樣的：
畢竟我們這個森林是需要合并的，當合并的時候呢，就要把unAddedMap里的那個結點換掉臨時結點（未來的老祖宗——新根）的某個兒子（我WA的那次就是因為這個原因，太難了），然后在unAddedMap里遞歸刪除，在map里遞歸插入。。。。

這需要兩個static的輔助方法（private即可），很nice！！！

遞歸刪除：

private static void deleteRecursively(Node root, Map<Character, Node> unAddedMap) {if (root == null) {return;}unAddedMap.remove(root.data);deleteRecursively(root.left, unAddedMap);deleteRecursively(root.right, unAddedMap);}

遞歸插入：

private static void addRecursively(Node root, Map<Character, Node> map) {if (root == null) {return;}map.put(root.data, root);addRecursively(root.left, map);addRecursively(root.right, map);}

最后的結果也需要遞歸先序遍歷：

private static void preOrder(Node root) {if (root == null) {return;}System.out.print(root.data);preOrder(root.left);preOrder(root.right);}

第一次提交——CE

這次CE就是因為沒把Node類作為內部類，不能被OJ識別所以判錯。。。。

第二次提交——WA

這個WA就是因為沒換，具體原因我忘了誒，代碼奉上，可以自己發掘其“魅力”（毫無魅力的破代碼）
這是錯的哈，正解在下面。。。

import java.util.*;public class Main {static class Node {char data;Node left, right;Node(char data) {this.data = data;this.left = null;this.right = null;}@Overridepublic boolean equals(Object o) {if (this == o) return true;if (o == null || getClass() != o.getClass()) return false;Node node = (Node) o;return data == node.data;}@Overridepublic int hashCode() {return Objects.hash(data);}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int num = Integer.parseInt(scanner.nextLine());if (num == 0) {return;}//樹非空，進行一系列初始化Map<Character, Node> map = new HashMap<>(num);Map<Character, Node> unAddedMap = new HashMap<>(num);char[] firstChars = scanner.nextLine().toCharArray();//初始化根結點Node root = new Node(firstChars[0]);map.put(firstChars[0], root);if (firstChars[1] != '*') {root.left = new Node(firstChars[1]);map.put(firstChars[1], root.left);}if (firstChars[2] != '*') {root.right = new Node(firstChars[2]);map.put(firstChars[2], root.right);}//循環num-1次for (int i = 1; i < num; i++) {char[] chars = scanner.nextLine().toCharArray();//根、左兒子、右兒子的charchar tempRootChar = chars[0], tempLeftChar = chars[1], tempRightChar = chars[2];//臨時的根結點、左兒子結點、右兒子結點Node tempRoot, tempLeft, tempRight;//Map里已有，也就是說已經存在臨時父結點，而且子結點也有咯if (map.containsKey(tempRootChar)) {tempRoot = map.get(tempRootChar);//子結點必定不存在于Map，但要看看unAddedif (unAddedMap.containsKey(tempLeftChar)) {tempLeft = unAddedMap.get(tempLeftChar);root.left = tempLeft;addRecursively(tempLeft, map);deleteRecursively(tempLeft, unAddedMap);} else if (tempLeftChar == '*') {tempLeft = null;} else {tempLeft = new Node(tempLeftChar);}if (unAddedMap.containsKey(tempRightChar)) {tempRight = unAddedMap.get(tempRightChar);root.right = tempRight;addRecursively(tempRight, map);deleteRecursively(tempRight, unAddedMap);} else if (tempRightChar == '*') {tempRight = null;} else {tempRight = new Node(tempRightChar);}//插入左右兒子節點tempRoot.left = tempLeft;tempRoot.right = tempRight;} else if (unAddedMap.containsKey(tempRootChar)) { //這就是說它的父結點已經存在咯，然后還沒插進樹里//繼續放在unAdded里tempRoot = unAddedMap.get(tempRootChar);//子結點必定不存在//插入左右兒子節點if (tempLeftChar != '*') {tempLeft = new Node(tempLeftChar);tempRoot.left = tempLeft;unAddedMap.put(tempLeftChar, tempLeft);}if (tempRightChar != '*') {tempRight = new Node(tempRightChar);tempRoot.right = tempRight;unAddedMap.put(tempRightChar, tempRight);}} else { //兩邊都沒有，看看是不是子結點就是現在的根tempRoot = new Node(tempRootChar);if (tempLeftChar == '*') {tempLeft = null;} else {tempLeft = new Node(tempLeftChar);tempRoot.left = tempLeft;}if (tempRightChar == '*') {tempRight = null;} else {tempRight = new Node(tempRightChar);tempRoot.right = tempRight;}if (root.equals(tempLeft)) { //左兒子是根結點，換根tempLeft = root;tempRoot.left = tempLeft;root = tempRoot;//查找右兒子if (unAddedMap.containsKey(tempRightChar)) {tempRight = unAddedMap.get(tempRightChar);root.right = tempRight;addRecursively(tempRight, map);deleteRecursively(tempRight, unAddedMap);}} else if (root.equals(tempRight)) { //右兒子是根結點，換根tempRight = root;tempRoot.right = tempRight;root = tempRoot;//查找左兒子if (unAddedMap.containsKey(tempLeftChar)) {tempLeft = unAddedMap.get(tempLeftChar);root.left = tempLeft;addRecursively(tempLeft, map);deleteRecursively(tempLeft, unAddedMap);}} else { //插入unAdded里if (unAddedMap.containsKey(tempLeftChar)) {tempRoot.left = unAddedMap.get(tempLeftChar);} else if (tempLeftChar != '*') {unAddedMap.put(tempLeftChar, tempLeft);}if (unAddedMap.containsKey(tempRightChar)) {tempRoot.right = unAddedMap.get(tempRightChar);} else if (tempRightChar != '*') {unAddedMap.put(tempRightChar, tempRight);}unAddedMap.put(tempRootChar, tempRoot);}}}scanner.close();preOrder(root);}private static void preOrder(Node root) {if (root == null) {return;}System.out.print(root.data);preOrder(root.left);preOrder(root.right);}private static void deleteRecursively(Node root, Map<Character, Node> unAddedMap) {if (root == null) {return;}unAddedMap.remove(root.data);deleteRecursively(root.left, unAddedMap);deleteRecursively(root.right, unAddedMap);}private static void addRecursively(Node root, Map<Character, Node> map) {if (root == null) {return;}map.put(root.data, root);addRecursively(root.left, map);addRecursively(root.right, map);}}

哦~~我竟截圖為證。。。
好吧，改的是這里，就AC了：

AC代碼（Java語言描述）

import java.util.*;public class Main {static class Node {char data;Node left, right;Node(char data) {this.data = data;this.left = null;this.right = null;}@Overridepublic boolean equals(Object o) {if (this == o) return true;if (o == null || getClass() != o.getClass()) return false;Node node = (Node) o;return data == node.data;}@Overridepublic int hashCode() {return Objects.hash(data);}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int num = Integer.parseInt(scanner.nextLine());if (num == 0) {return;}//樹非空，進行一系列初始化Map<Character, Node> map = new HashMap<>(num);Map<Character, Node> unAddedMap = new HashMap<>(num);char[] firstChars = scanner.nextLine().toCharArray();//初始化根結點Node root = new Node(firstChars[0]);map.put(firstChars[0], root);if (firstChars[1] != '*') {root.left = new Node(firstChars[1]);map.put(firstChars[1], root.left);}if (firstChars[2] != '*') {root.right = new Node(firstChars[2]);map.put(firstChars[2], root.right);}//循環num-1次for (int i = 1; i < num; i++) {char[] chars = scanner.nextLine().toCharArray();//根、左兒子、右兒子的charchar tempRootChar = chars[0], tempLeftChar = chars[1], tempRightChar = chars[2];//臨時的根結點、左兒子結點、右兒子結點Node tempRoot, tempLeft, tempRight;//Map里已有，也就是說已經存在臨時父結點，而且子結點肯定沒有if (map.containsKey(tempRootChar)) {tempRoot = map.get(tempRootChar);//子結點必定不存在于Map，但要看看unAddedif (unAddedMap.containsKey(tempLeftChar)) {tempLeft = unAddedMap.get(tempLeftChar);root.left = tempLeft;addRecursively(tempLeft, map);deleteRecursively(tempLeft, unAddedMap);} else if (tempLeftChar == '*') {tempLeft = null;} else {tempLeft = new Node(tempLeftChar);map.put(tempLeftChar, tempLeft);}if (unAddedMap.containsKey(tempRightChar)) {tempRight = unAddedMap.get(tempRightChar);root.right = tempRight;addRecursively(tempRight, map);deleteRecursively(tempRight, unAddedMap);} else if (tempRightChar == '*') {tempRight = null;} else {tempRight = new Node(tempRightChar);map.put(tempRightChar, tempRight);}//插入左右兒子節點tempRoot.left = tempLeft;tempRoot.right = tempRight;} else if (unAddedMap.containsKey(tempRootChar)) { //這就是說它的父結點已經存在咯，然后還沒插進樹里//繼續放在unAdded里tempRoot = unAddedMap.get(tempRootChar);//子結點必定不存在//插入左右兒子節點if (tempLeftChar != '*') {tempLeft = new Node(tempLeftChar);tempRoot.left = tempLeft;unAddedMap.put(tempLeftChar, tempLeft);}if (tempRightChar != '*') {tempRight = new Node(tempRightChar);tempRoot.right = tempRight;unAddedMap.put(tempRightChar, tempRight);}} else { //兩邊都沒有，看看是不是子結點就是現在的根tempRoot = new Node(tempRootChar);if (tempLeftChar == '*') {tempLeft = null;} else {tempLeft = new Node(tempLeftChar);tempRoot.left = tempLeft;}if (tempRightChar == '*') {tempRight = null;} else {tempRight = new Node(tempRightChar);tempRoot.right = tempRight;}if (root.equals(tempLeft)) { //左兒子是根結點，換根tempLeft = root;tempRoot.left = tempLeft;root = tempRoot;//查找右兒子if (unAddedMap.containsKey(tempRightChar)) {tempRight = unAddedMap.get(tempRightChar);root.right = tempRight;addRecursively(tempRight, map);deleteRecursively(tempRight, unAddedMap);}} else if (root.equals(tempRight)) { //右兒子是根結點，換根tempRight = root;tempRoot.right = tempRight;root = tempRoot;//查找左兒子if (unAddedMap.containsKey(tempLeftChar)) {tempLeft = unAddedMap.get(tempLeftChar);root.left = tempLeft;addRecursively(tempLeft, map);deleteRecursively(tempLeft, unAddedMap);}} else { //插入unAdded里if (unAddedMap.containsKey(tempLeftChar)) {tempRoot.left = unAddedMap.get(tempLeftChar);} else if (tempLeftChar != '*') {unAddedMap.put(tempLeftChar, tempLeft);}if (unAddedMap.containsKey(tempRightChar)) {tempRoot.right = unAddedMap.get(tempRightChar);} else if (tempRightChar != '*') {unAddedMap.put(tempRightChar, tempRight);}unAddedMap.put(tempRootChar, tempRoot);}}}scanner.close();preOrder(root);}private static void preOrder(Node root) {if (root == null) {return;}System.out.print(root.data);preOrder(root.left);preOrder(root.right);}private static void deleteRecursively(Node root, Map<Character, Node> unAddedMap) {if (root == null) {return;}unAddedMap.remove(root.data);deleteRecursively(root.left, unAddedMap);deleteRecursively(root.right, unAddedMap);}private static void addRecursively(Node root, Map<Character, Node> map) {if (root == null) {return;}map.put(root.data, root);addRecursively(root.left, map);addRecursively(root.right, map);}}

感想

這代碼真簡（la）潔（ji）。。。
我迷了，不過自己手寫一次滿有收獲的誒，奧利給！！！！
容我休息一下~~~
寫代碼好快（tou）樂（tu）。。。。
您能看完，是賞臉的，菜雞拜謝Orz

傳說七只Orz的自己可以召喚神龍哇！！！

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