[英]N-Ary Tree C++ - How to find the level of a node
I would like to returns the level of a given node. 我想返回给定节点的级别。 I've been able to do this for binary trees but for n-ary trees there is no way to run it.
我已经能够对二叉树执行此操作,但对于n元树则无法运行它。 Any ideas ?
有任何想法吗 ?
For the binary tree the solution was: 对于二叉树,解决方案是:
int findLevel(BinAlbero<int>::node root, BinAlbero<int>::node ptr,
int level = 0) {
if (root == NULL)
return -1;
if (root == ptr)
return level;
// If NULL or leaf Node
if (root->left == NULL && root->right == NULL)
return -1;
// Find If ptr is present in the left or right subtree.
int levelLeft = findLevel(root->left, ptr, level + 1);
int levelRight = findLevel(root->right, ptr, level + 1);
if (levelLeft == -1)
return levelRight;
else
return levelLeft;}
where "ptr" is the node for which the level is searched. 其中“ ptr”是要搜索其级别的节点。 Thank you.
谢谢。 Here is the structure of N-Ary Tree:
这是N-Ary树的结构:
class AlberoN {
public:
typedef T tipoelem;
typedef bool boolean;
struct nodoAlbero {
tipoelem elemento;
struct nodoAlbero* parent;
/*Primo figlio*/
struct nodoAlbero* children;
struct nodoAlbero* brother;
};
typedef nodoAlbero* node;
/*......*/
private:
nodo root;};
If i use this tree: 如果我使用这棵树:
8
/ / \ \
17 30 18 7
/
15
/ \
51 37
I tried but the function returns the exact level only for node 17 and 15. With this code: 我尝试过,但是该函数仅返回节点17和15的确切级别。使用以下代码:
int findLevel(AlberoN<int> t, AlberoN<int>::nodo root, AlberoN<int>::nodo ptr,
int level = 0) {
if (root == ptr) {
return level;}
if (root == NULL)
return -1;
if (!t.leaf(root)) {
level++;
root = t.firstSon(root);
findLevel(t, root, ptr, level);}
if (!t.lastBrother(root)) {
root = t.succBrother(root);
findLevel(t, root, ptr, level);}
return level;}
int livellofiglio = findLevel(temp, ptr, level + 1);
while (temp != NULL) {
temp = t.succBrother(temp);
int livellofratello = findLevel(temp, ptr, level + 1);
if (livellofiglio == -1)
return livellofratello;
else
return livellofiglio;
}
You will always return after a single iteration of your loop, so you only ever visit the first two child nodes of a given node. 在循环的一次迭代之后,您将始终返回,因此您只能访问给定节点的前两个子节点。
You should always be iterating over the entire array, and return the found value (if present): 您应该始终遍历整个数组,并返回找到的值(如果存在):
while (temp != NULL) {
int livellofratello = findLevel(temp, ptr, level + 1);
if (livellofratello != -1)
return livellofratello;
temp = t.succBrother(temp);
}
return -1
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