#practiceLinkDiv { display: none! importante; }Dato un albero binario trovare la lunghezza del percorso più lungo che comprende nodi con valori consecutivi in ordine crescente. Ogni nodo è considerato come un percorso di lunghezza 1.
Esempi:
In below diagram binary tree with longest consecutive path(LCP) are shown :
Possiamo risolvere il problema sopra in modo ricorsivo. In ogni nodo abbiamo bisogno delle informazioni del suo nodo genitore, se il nodo corrente ha un valore in più rispetto al suo nodo genitore, allora crea un percorso consecutivo in ciascun nodo, confronteremo il valore del nodo con il suo valore genitore e aggiorneremo di conseguenza il percorso consecutivo più lungo.
Per ottenere il valore del nodo genitore passeremo (node_value + 1) come argomento al metodo ricorsivo e confronteremo il valore del nodo con questo valore dell'argomento se soddisfa aggiorna la lunghezza corrente del percorso consecutivo altrimenti reinizializza la lunghezza del percorso corrente di 1.
Si prega di consultare il codice seguente per una migliore comprensione:
C++// C/C++ program to find longest consecutive // sequence in binary tree #include
// 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 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
<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>
Produzione
3
Complessità temporale: O(N) dove N è il numero di nodi in un dato albero binario.
Spazio ausiliario: O(log(N))
Discusso anche al link seguente:
Lunghezza massima del percorso crescente consecutivo nell'albero binario