C++中N-Ary树的深度？
让我们首先定义表示包含字符键和Node*向量的树节点的结构。
struct Node{
char key;
vector children;
}; 接下来我们创建我们的createNode(intkey)函数，它接受一个int键值并将其分配给节点的键成员。该函数返回指向创建的结构节点的指针。
Node *createNode(int key){
Node *node = new Node;
node->key = key;
return node;
}我们的depthOfTree(structNode*root)函数将根节点作为参数。如果根为NULL，则深度返回为0。
int depthOfTree(struct Node *root){
if (root==NULL)
return 0;然后我们定义maxDepth变量并将其值赋值为0。然后我们遍历树的所有子节点并在每次递归时增加maxDepth。一旦满足基本条件并且递归结束，我们就会返回maxDepth。
int depthOfTree(struct Node *root){
if (root==NULL)
return 0;
int maxDepth = 0;
for(auto i: root->children){
maxDepth = depthOfTree(i) + 1;
}
return maxDepth;
}示例
让我们看看下面的实现来找到N-Ary树的深度-
#include输出结果#include using namespace std; struct Node{ char key; vector children; }; Node *createNode(int key){ Node *node = new Node; node->key = key; return node; } int depthOfTree(struct Node *root){ if (root==NULL) return 0; int maxDepth = 0; for(auto i: root->children){ maxDepth = depthOfTree(i) + 1; } return maxDepth; } int main(){ Node *root = createNode('S'); (root->children).push_back(createNode('O')); (root->children).push_back(createNode('A')); (root->children).push_back(createNode('D')); (root->children).push_back(createNode('N')); (root->children[0]->children).push_back(createNode('L')); (root->children[0]->children).push_back(createNode('I')); (root->children[2]->children).push_back(createNode('R')); (root->children[3]->children).push_back(createNode('C')); (root->children[3]->children).push_back(createNode('H')); (root->children[3]->children).push_back(createNode('I')); cout <<"The depth of the n-ary tree is "<< depthOfTree(root) << endl; return 0; }
上面的代码将产生以下输出-
The depth of the n-ary tree is 2