1600. Throne Inheritance

# 1600. 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`.

## 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 #

``````package leetcode

type ThroneInheritance struct {
king  string
edges map[string][]string
}

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) {
}

func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
var preorder func(string)
preorder = func(name string) {
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();
*/
``````

Nov 25, 2022