# [1600. Throne Inheritance](https://leetcode.com/problems/throne-inheritance/) ## 题目 A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born. The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let's define the recursive function `Successor(x, curOrder)`, which given a person `x` and the inheritance order so far, returns who should be the next person after `x` in the order of inheritance. ``` Successor(x, curOrder): if x has no children or all of x's children are in curOrder: if x is the king return null else return Successor(x's parent, curOrder) else return x's oldest child who's not in curOrder ``` For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice's son Jack. 1. In the beginning, `curOrder` will be `["king"]`. 2. Calling `Successor(king, curOrder)` will return Alice, so we append to `curOrder` to get `["king", "Alice"]`. 3. Calling `Successor(Alice, curOrder)` will return Jack, so we append to `curOrder` to get `["king", "Alice", "Jack"]`. 4. Calling `Successor(Jack, curOrder)` will return Bob, so we append to `curOrder` to get `["king", "Alice", "Jack", "Bob"]`. 5. Calling `Successor(Bob, curOrder)` will return `null`. Thus the order of inheritance will be `["king", "Alice", "Jack", "Bob"]`. Using the above function, we can always obtain a unique order of inheritance. Implement the `ThroneInheritance` class: - `ThroneInheritance(string kingName)` Initializes an object of the `ThroneInheritance` class. The name of the king is given as part of the constructor. - `void birth(string parentName, string childName)` Indicates that `parentName` gave birth to `childName`. - `void death(string name)` Indicates the death of `name`. The death of the person doesn't affect the `Successor` function nor the current inheritance order. You can treat it as just marking the person as dead. - `string[] getInheritanceOrder()` Returns a list representing the current order of inheritance **excluding** dead people. **Example 1:** ``` Input ["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"] [["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]] Output [null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]] Explanation ThroneInheritance t= new ThroneInheritance("king"); // order:king t.birth("king", "andy"); // order: king >andy t.birth("king", "bob"); // order: king > andy >bob t.birth("king", "catherine"); // order: king > andy > bob >catherine t.birth("andy", "matthew"); // order: king > andy >matthew > bob > catherine t.birth("bob", "alex"); // order: king > andy > matthew > bob >alex > catherine t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex >asha > catherine t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"] t.death("bob"); // order: king > andy > matthew >bob > alex > asha > catherine t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"] ``` **Constraints:** - `1 <= kingName.length, parentName.length, childName.length, name.length <= 15` - `kingName`, `parentName`, `childName`, and `name` consist of lowercase English letters only. - All arguments `childName` and `kingName` are **distinct**. - All `name` arguments of `death` will be passed to either the constructor or as `childName` to `birth` first. - For each call to `birth(parentName, childName)`, it is guaranteed that `parentName` is alive. - At most `105` calls will be made to `birth` and `death`. - At most `10` calls will be made to `getInheritanceOrder`. ## 题目大意 一个王国里住着国王、他的孩子们、他的孙子们等等。每一个时间点,这个家庭里有人出生也有人死亡。这个王国有一个明确规定的皇位继承顺序,第一继承人总是国王自己。我们定义递归函数 Successor(x, curOrder) ,给定一个人 x 和当前的继承顺序,该函数返回 x 的下一继承人。 ## 解题思路 - 这道题思路不难。先将国王每个孩子按照顺序存在一个 map 中,然后每个国王的孩子还存在父子关系,同理也按顺序存在 map 中。执行 GetInheritanceOrder() 函数时,将国王的孩子按顺序遍历,如果每个孩子还有孩子,递归遍历到底。如果把继承关系看成一棵树,此题便是多叉树的先根遍历的问题。 ## 代码 ```go package leetcode type ThroneInheritance struct { king string edges map[string][]string dead map[string]bool } func Constructor(kingName string) (t ThroneInheritance) { return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}} } func (t *ThroneInheritance) Birth(parentName, childName string) { t.edges[parentName] = append(t.edges[parentName], childName) } func (t *ThroneInheritance) Death(name string) { t.dead[name] = true } func (t *ThroneInheritance) GetInheritanceOrder() (res []string) { var preorder func(string) preorder = func(name string) { if !t.dead[name] { res = append(res, name) } for _, childName := range t.edges[name] { preorder(childName) } } preorder(t.king) return } /** * Your ThroneInheritance object will be instantiated and called as such: * obj := Constructor(kingName); * obj.Birth(parentName,childName); * obj.Death(name); * param_3 := obj.GetInheritanceOrder(); */ ```