使用C++编写，找到N叉树中给定节点的兄弟节点数量

在本文中，我们将提供完整的信息来确定 n 叉树中给定节点的兄弟节点数量。我们需要使用用户给定的 key 值找到该节点的兄弟节点；如果不是，则输出-1。我们只能使用一种方法 -

简单方法

在这种方法中，我们将遍历所有节点并检查子节点是否与用户具有相同的值。如果存在，我们回答子节点的数量 - 1（给定值）。

示例

#include using namespace std; class Node { // structure of nodes of our tree. public: int key; vector child; Node(int data){ key = data; } }; int main(){ // Building The Tree Node* Base = new Node(50); (Base->child).push_back(new Node(2)); (Base->child).push_back(new Node(30)); (Base->child).push_back(new Node(14)); (Base->child).push_back(new Node(60)); (Base->child[0]->child).push_back(new Node(15)); (Base->child[0]->child).push_back(new Node(25)); (Base->child[0]->child[1]->child).push_back(new Node(70)); (Base->child[0]->child[1]->child).push_back(new Node(100)); (Base->child[1]->child).push_back(new Node(6)); (Base->child[1]->child).push_back(new Node(1)); (Base->child[2]->child).push_back(new Node(7)); (Base->child[2]->child[0]->child).push_back(new Node(17)); (Base->child[2]->child[0]->child).push_back(new Node(99)); (Base->child[2]->child[0]->child).push_back(new Node(27)); (Base->child[3]->child).push_back(new Node(16)); int x = 30; queue q; q.push(Base); bool flag = 0; int answer = -1; if(Base -> key != x){ while(!q.empty()){ auto parent = q.front(); q.pop(); for(int i = 0; i child.size(); i++){ if(parent -> child[i] -> key == x){ answer = parent -> child.size() - 1; flag = 1; break; } q.push(parent -> child[i]); } if(flag) break; } cout