DUBINA N-ARY STABLA

S obzirom na n-arno stablo koji sadrži pozitivne vrijednosti čvora zadatak je pronaći dubina od stabla.
Bilješka: An n-arno stablo je stablo gdje svaki čvor može imati nula ili više dječji čvorovi. Za razliku od binarnog stabla koje ima najviše dva djeteta po čvoru (lijevo i desno) n-arno stablo dopušta više grana ili djeca za svaki čvor.

Primjeri:

Ulazni:

Izlaz: 3
Obrazloženje: Najduži put od korijena (čvor 81) do lista je ili 81 -> 26 -> 95 ili 81 -> 26 -> 86 što daje najveću dubinu od 3.
Ulazni:

Izlaz: 2
Obrazloženje: Najduži put od korijena (čvor 4) do bilo kojeg lista (čvorovi 5 ili 7) je 2 jer zahtijeva samo dvije razine obilaženja.
naučiti selen

Pristup:

Ideja je izračunati dubina N-arnog stabla rekurzivno inicijalizirati kao 0 zatim rekurzivno izračunajte dubina za svako dijete i pratiti najveća dubina naišao. Na kraju dodajte 1 do maksimalne dubine (za trenutni čvor) i vratite proizlaziti . Ovaj pristup osigurava da pronađemo najduži put od korijena do bilo kojeg lisnog čvora.

N-Ary stablo se može obilaziti baš kao i normalno stablo. Samo moramo uzeti u obzir sve potomke određenog čvora i rekurzivno pozvati tu funkciju na svakom čvoru.

C++

// C++ Code to find the depth of an N-ary tree #include    using namespace std; class Node { public:  int data;  vector<Node*> children;  Node(int val) {  data = val;  } }; // Recursive function to calculate maximum depth int maxDepth(Node* root) {    // If the node is null depth is 0  if (!root) {  return 0;  }  int depth = 0;  // Recur for all children and find the maximum depth  for (auto child : root->children) {  depth = max(depth maxDepth(child));  }  // Add 1 to include the current node  // in the depth count  return depth + 1; } int main() {  // Representation of given N-ary tree  // 1  // / |   // 2 3 4  // /   // 5 6  Node* root = new Node(1);  root->children.push_back(new Node(2));  root->children.push_back(new Node(3));  root->children.push_back(new Node(4));  root->children[0]->children.push_back(new Node(5));  root->children[2]->children.push_back(new Node(6));  cout << maxDepth(root);  return 0; }

Java

// Java Code to find the depth of an N-ary tree import java.util.*; class Node {  int data;  List<Node> children;  Node(int val) {  data = val;  children = new ArrayList<>();  } } // Recursive function to calculate // maximum depth class GfG {    static int maxDepth(Node root) {  // If the node is null depth is 0  if (root == null) {  return 0;  }  int depth = 0;  // Recur for all children and find  // the maximum depth  for (Node child : root.children) {  depth = Math.max(depth maxDepth(child));  }  // Add 1 to include the current node   // in the depth count  return depth + 1;  }  public static void main(String[] args) {  // Representation of given N-ary tree  // 1  // / |   // 2 3 4  // /   // 5 6  Node root = new Node(1);  root.children.add(new Node(2));  root.children.add(new Node(3));  root.children.add(new Node(4));  root.children.get(0).children.add(new Node(5));  root.children.get(2).children.add(new Node(6));  System.out.println(maxDepth(root));  } }

Python

# Python Code to find the depth  # of an N-ary tree class Node: def __init__(self val): self.data = val self.children = [] # Recursive function to calculate # maximum depth def max_depth(root): # If the node is None depth is 0 if not root: return 0 depth = 0 # Recur for all children and  # find the maximum depth for child in root.children: depth = max(depth max_depth(child)) # Add 1 to include the current # node in the depth count return depth + 1 if __name__ == '__main__': # Representation of given N-ary tree # 1 # / |  # 2 3 4 # /  # 5 6 root = Node(1) root.children.append(Node(2)) root.children.append(Node(3)) root.children.append(Node(4)) root.children[0].children.append(Node(5)) root.children[2].children.append(Node(6)) print(max_depth(root))

// C# Code to find the depth of an N-ary tree using System; using System.Collections.Generic; class Node {  public int data;  public List<Node> children;  public Node(int val) {  data = val;  children = new List<Node>();  } } // Recursive function to calculate // maximum depth class GfG {    static int MaxDepth(Node root) {  // If the node is null depth is 0  if (root == null) {  return 0;  }  int depth = 0;  // Recur for all children and find the maximum depth  foreach (Node child in root.children) {  depth = Math.Max(depth MaxDepth(child));  }  // Add 1 to include the current  // node in the depth count  return depth + 1;  }  static void Main(string[] args) {  // Representation of given N-ary tree  // 1  // / |   // 2 3 4  // /   // 5 6  Node root = new Node(1);  root.children.Add(new Node(2));  root.children.Add(new Node(3));  root.children.Add(new Node(4));  root.children[0].children.Add(new Node(5));  root.children[2].children.Add(new Node(6));  Console.WriteLine(MaxDepth(root));  } }

JavaScript

// JavaScript Code to find the depth  // of an N-ary tree class Node {  constructor(val) {  this.data = val;  this.children = [];  } } // Recursive function to calculate  // maximum depth function maxDepth(root) {  // If the node is null depth is 0  if (!root) {  return 0;  }  let depth = 0;  // Recur for all children and find  // the maximum depth  for (let child of root.children) {  depth = Math.max(depth maxDepth(child));  }  // Add 1 to include the current node   // in the depth count  return depth + 1; } // Representation of given N-ary tree // 1 // / |  // 2 3 4 // /  // 5 6 const root = new Node(1); root.children.push(new Node(2)); root.children.push(new Node(3)); root.children.push(new Node(4)); root.children[0].children.push(new Node(5)); root.children[2].children.push(new Node(6)); console.log(maxDepth(root));

Izlaz

Vremenska složenost: O(n) budući da se svaki čvor posjećuje jednom, gdje je n ukupan broj čvorova u N-arnom stablu.
Pomoćni prostor: O(h) gdje je h visina stabla zbog korištenja rekurzivnog stoga poziva.

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Dubina N-Ary stabla