LANGSTE OPEENVOLGENDE REEKS IN BINAIRE BOOM

Probeer het eens op GfG Practice Langste opeenvolgende reeks in binaire boom' title=

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Gegeven een binaire boom, zoek de lengte van het langste pad dat bestaat uit knooppunten met opeenvolgende waarden in oplopende volgorde. Elk knooppunt wordt beschouwd als een pad met lengte 1.
Voorbeelden:

In below diagram binary tree with longest consecutive path(LCP) are shown :

Java lokale datum

Aanbevolen praktijk Langste opeenvolgende reeks in binaire boom Probeer het!

We kunnen het bovenstaande probleem recursief oplossen. Bij elk knooppunt hebben we informatie nodig over het bovenliggende knooppunt. Als het huidige knooppunt een waarde heeft die één hoger is dan het bovenliggende knooppunt, dan maakt het een opeenvolgend pad. Bij elk knooppunt vergelijken we de waarde van het knooppunt met de bovenliggende waarde en werken we het langste opeenvolgende pad dienovereenkomstig bij.

Om de waarde van het bovenliggende knooppunt te verkrijgen, geven we de (node_value + 1) als argument door aan de recursieve methode en vergelijken we de knooppuntwaarde met deze argumentwaarde als deze voldoet, update de huidige lengte van het opeenvolgende pad, anders initialiseren we de huidige padlengte opnieuw met 1.

Zie onderstaande code voor een beter begrip:

C++

// C/C++ program to find longest consecutive // sequence in binary tree #include    using namespace std; /* A binary tree node has data pointer to left  child and a pointer to right child */ struct Node {  int data;  Node *left *right; }; // A utility function to create a node Node* newNode(int data) {  Node* temp = new Node;  temp->data = data;  temp->left = temp->right = NULL;  return temp; } // Utility method to return length of longest // consecutive sequence of tree void longestConsecutiveUtil(Node* root int curLength  int expected int& res) {  if (root == NULL)  return;  // if root data has one more than its parent  // then increase current length  if (root->data == expected)  curLength++;  else  curLength = 1;  // update the maximum by current length  res = max(res curLength);  // recursively call left and right subtree with  // expected value 1 more than root data  longestConsecutiveUtil(root->left curLength  root->data + 1 res);  longestConsecutiveUtil(root->right curLength  root->data + 1 res); } // method returns length of longest consecutive // sequence rooted at node root int longestConsecutive(Node* root) {  if (root == NULL)  return 0;  int res = 0;  // call utility method with current length 0  longestConsecutiveUtil(root 0 root->data res);  return res; } // Driver code to test above methods int main() {  Node* root = newNode(6);  root->right = newNode(9);  root->right->left = newNode(7);  root->right->right = newNode(10);  root->right->right->right = newNode(11);  printf('%dn' longestConsecutive(root));  return 0; }

Java

// Java program to find longest consecutive  // sequence in binary tree class Node {  int data;  Node left right;  Node(int item)  {  data = item;  left = right = null;  } } class Result  {  int res = 0; } class BinaryTree {  Node root;  // method returns length of longest consecutive   // sequence rooted at node root   int longestConsecutive(Node root)  {  if (root == null)  return 0;  Result res = new Result();    // call utility method with current length 0   longestConsecutiveUtil(root 0 root.data res);    return res.res;  }  // Utility method to return length of longest   // consecutive sequence of tree   private void longestConsecutiveUtil(Node root int curlength   int expected Result res)  {  if (root == null)  return;  // if root data has one more than its parent   // then increase current length   if (root.data == expected)  curlength++;  else  curlength = 1;  // update the maximum by current length   res.res = Math.max(res.res curlength);  // recursively call left and right subtree with   // expected value 1 more than root data   longestConsecutiveUtil(root.left curlength root.data + 1 res);  longestConsecutiveUtil(root.right curlength root.data + 1 res);  }  // Driver code  public static void main(String args[])   {  BinaryTree tree = new BinaryTree();  tree.root = new Node(6);  tree.root.right = new Node(9);  tree.root.right.left = new Node(7);  tree.root.right.right = new Node(10);  tree.root.right.right.right = new Node(11);  System.out.println(tree.longestConsecutive(tree.root));  } } // This code is contributed by shubham96301

Python3

# Python3 program to find longest consecutive  # sequence in binary tree  # A utility class to create a node  class newNode: def __init__(self data): self.data = data self.left = self.right = None # Utility method to return length of  # longest consecutive sequence of tree  def longestConsecutiveUtil(root curLength expected res): if (root == None): return # if root data has one more than its  # parent then increase current length  if (root.data == expected): curLength += 1 else: curLength = 1 # update the maximum by current length  res[0] = max(res[0] curLength) # recursively call left and right subtree  # with expected value 1 more than root data  longestConsecutiveUtil(root.left curLength root.data + 1 res) longestConsecutiveUtil(root.right curLength root.data + 1 res) # method returns length of longest consecutive  # sequence rooted at node root  def longestConsecutive(root): if (root == None): return 0 res = [0] # call utility method with current length 0  longestConsecutiveUtil(root 0 root.data res) return res[0] # Driver Code if __name__ == '__main__': root = newNode(6) root.right = newNode(9) root.right.left = newNode(7) root.right.right = newNode(10) root.right.right.right = newNode(11) print(longestConsecutive(root)) # This code is contributed by PranchalK

// C# program to find longest consecutive  // sequence in binary tree  using System;  class Node  {   public int data;   public Node left right;   public Node(int item)   {   data = item;   left = right = null;   }  }  class Result  {   public int res = 0;  }  class GFG  {   Node root;   // method returns length of longest consecutive   // sequence rooted at node root   int longestConsecutive(Node root)   {   if (root == null)   return 0;   Result res = new Result();     // call utility method with current length 0   longestConsecutiveUtil(root 0 root.data res);     return res.res;   }   // Utility method to return length of longest   // consecutive sequence of tree   private void longestConsecutiveUtil(Node root int curlength   int expected Result res)   {   if (root == null)   return;   // if root data has one more than its parent   // then increase current length   if (root.data == expected)   curlength++;   else  curlength = 1;   // update the maximum by current length   res.res = Math.Max(res.res curlength);   // recursively call left and right subtree with   // expected value 1 more than root data   longestConsecutiveUtil(root.left curlength   root.data + 1 res);   longestConsecutiveUtil(root.right curlength   root.data + 1 res);   }   // Driver code   public static void Main(String []args)   {   GFG tree = new GFG();   tree.root = new Node(6);   tree.root.right = new Node(9);   tree.root.right.left = new Node(7);   tree.root.right.right = new Node(10);   tree.root.right.right.right = new Node(11);   Console.WriteLine(tree.longestConsecutive(tree.root));   }  }  // This code is contributed by 29AjayKumar

JavaScript

<script> // JavaScript program to find longest consecutive  // sequence in binary tree class Node {  constructor(item)  {  this.data=item;  this.left = this.right = null;  } } let res = 0; let root; function longestConsecutive(root) {  if (root == null)  return 0;    res=[0];     // call utility method with current length 0   longestConsecutiveUtil(root 0 root.data res);    return res[0]; }  // Utility method to return length of longest   // consecutive sequence of tree  function longestConsecutiveUtil(rootcurlength expectedres) {  if (root == null)  return;    // if root data has one more than its parent   // then increase current length   if (root.data == expected)  curlength++;  else  curlength = 1;    // update the maximum by current length   res[0] = Math.max(res[0] curlength);    // recursively call left and right subtree with   // expected value 1 more than root data   longestConsecutiveUtil(root.left curlength   root.data + 1 res);  longestConsecutiveUtil(root.right curlength   root.data + 1 res); } // Driver code root = new Node(6); root.right = new Node(9); root.right.left = new Node(7); root.right.right = new Node(10); root.right.right.right = new Node(11); document.write(longestConsecutive(root)); // This code is contributed by rag2127 </script>

Uitvoer

Tijdcomplexiteit: O(N) waarbij N het aantal knooppunten in een gegeven binaire boom is.
Hulpruimte: O(log(N))
Wordt ook besproken op onderstaande link:
Maximale opeenvolgende toenemende padlengte in binaire boom

staten in de VS