题目链接
一棵树，树中包含TRUE、FALSE、AND、OR四种结点，其中TRUE和FALSE是叶子结点，AND和OR结点的儿子包含多个结点，现在要求执行最少次数的以下操作：

• 把AND变成OR
• 把OR变成AND

使得整棵树的bool值结果翻转。

思路

只考虑当前结点和它的儿子。
要想改变当前结点的值，有两种方法：

• 改变当前结点的操作
• 改变当前结点的儿子结点的值

有时甚至需要把这两种方法结合起来，比如：当前结点为AND，它的儿子们全为TRUE。这是要改变当前结点的值，可能要把当前结点改为OR，并且把某个儿子改为FALSE。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;public class Main {
final int MAX_VLUE = 100000;class Node {List<Node> son = new ArrayList<>();String value;int parent;int reverseValue = -1;boolean nowValue = false;boolean inited = false;Node(String value, int parent) {this.value = value;this.parent = parent;}int reverseOne(boolean toValue) {int min = MAX_VLUE;for (Node i : son) {if (i.getValue() == toValue) return 0;min = Math.min(min, i.reverse());}return min;}int reverseAll(boolean toValue) {int s = 0;for (Node i : son) {if (i.getValue() != toValue) {s += i.reverse();}}return s;}//翻转需要的步数int reverse() {if (reverseValue != -1) return reverseValue;if (value.equals("TRUE") || value.equals("FALSE")) return MAX_VLUE;if (value.equals("AND")) {if (nowValue) {reverseValue = Math.min(reverseOne(false), 1 + reverseAll(false));} else {reverseValue = Math.min(reverseAll(true), 1 + reverseOne(true));}} else if (value.equals("OR")) {if (nowValue) {reverseValue = Math.min(1 + reverseOne(false), reverseAll(false));} else {reverseValue = Math.min(reverseOne(true), 1 + reverseAll(true));}}return reverseValue;}boolean getValue() {if (inited) return nowValue;inited = true;if (value.equals("TRUE")) {nowValue = true;return true;} else if (value.equals("FALSE")) {nowValue = false;return false;} else if (value.equals("AND")) {nowValue = son.stream().allMatch(Node::getValue);return nowValue;} else if (value.equals("OR")) {nowValue = son.stream().anyMatch(Node::getValue);return nowValue;}return false;}
}Main() {Scanner cin = new Scanner(System.in);int N = cin.nextInt();Node[] a = new Node[N + 1];a[0] = new Node("AND", 0);for (int i = 1; i <= N; i++) {int parent = cin.nextInt();String value = cin.next();a[i] = new Node(value, parent);}int root = 1;for (int i = 1; i <= N; i++) {a[a[i].parent].son.add(a[i]);if (a[i].parent == 0) {root = i;}}a[root].getValue();int step = a[root].reverse();if (step >= MAX_VLUE) {System.out.println(-1);} else {System.out.println(step);}
}public static void main(String[] args) {new Main();
}
}