#practiceLinkDiv { display: ingen !viktigt; }Givet ett binärt träd, hitta längden på den längsta vägen som består av noder med på varandra följande värden i ökande ordning. Varje nod betraktas som en väg med längd 1.
Exempel:
10 av 10
In below diagram binary tree with longest consecutive path(LCP) are shown :
Vi kan lösa ovanstående problem rekursivt. Vid varje nod behöver vi information om dess föräldernod om den nuvarande noden har ett värde ett mer än dess föräldernod, då gör den en på varandra följande väg vid varje nod, vi kommer att jämföra nodens värde med dess föräldervärde och uppdatera den längsta på varandra följande vägen därefter.
För att få värdet på den överordnade noden kommer vi att skicka (node_värde + 1) som ett argument till den rekursiva metoden och jämföra nodvärdet med detta argumentvärde, om det uppfyller, uppdatera den aktuella längden på den på varandra följande vägen, annars återinitiera den aktuella väglängden med 1.
java binära träd
Se nedanstående kod för bättre förståelse:
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>
Produktion
3
Tidskomplexitet: O(N) där N är antalet noder i ett givet binärt träd.
Hjälputrymme: O(log(N))
Diskuteras även på länken nedan:
Maximal konsekutiv ökande banlängd i binärt träd