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הרצף הארוך ביותר ברציפות בעץ בינארי

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בהינתן עץ בינארי מצא את אורך הנתיב הארוך ביותר הכולל צמתים עם ערכים עוקבים בסדר הולך וגדל. כל צומת נחשב כנתיב באורך 1.
דוגמאות:  
 

gimp כיצד לבטל את הבחירה
In below diagram binary tree with longest consecutive path(LCP) are shown :

תרגול מומלץ הרצף הארוך ביותר ברציפות בעץ בינארי נסה את זה!

אנחנו יכולים לפתור את הבעיה שלעיל באופן רקורסיבי. בכל צומת אנו זקוקים למידע על הצומת האב שלו אם לצומת הנוכחי יש ערך אחד יותר מהצומת האב שלו אז הוא עושה נתיב רצוף בכל צומת נשווה את הערך של הצומת עם ערך האב שלו ונעדכן את הנתיב הרצוף הארוך ביותר בהתאם. 



לקבלת הערך של צומת האב, נעביר את (node_value + 1) כארגומנט לשיטה הרקורסיבית ונשווה את ערך הצומת עם ערך הארגומנט הזה אם עונה על עדכון האורך הנוכחי של נתיב עוקב, אחרת אתחול מחדש את אורך הנתיב הנוכחי ב-1. 

אנא עיין בקוד למטה להבנה טובה יותר: 

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#
// 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> 

תְפוּקָה
3

מורכבות זמן: O(N) כאשר N הוא מספר הצמתים בעץ בינארי נתון.
רווח עזר: O(log(N))
נדון גם בקישור הבא: 
אורך נתיב מרבי רצוף הולך וגדל בעץ בינארי

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