Ermitteln Sie anhand eines Binärbaums die Länge des längsten Pfads, der aus Knoten mit aufeinanderfolgenden Werten in aufsteigender Reihenfolge besteht. Jeder Knoten wird als Pfad der Länge 1 betrachtet.
Beispiele:
10 / / 11 9 / / / / 13 12 13 8 Maximum Consecutive Path Length is 3 (10 11 12) Note : 10 9 8 is NOT considered since the nodes should be in increasing order. 5 / / 8 11 / / 9 10 / / / / 6 15 Maximum Consecutive Path Length is 2 (8 9).
Jeder Knoten im Binärbaum kann entweder Teil des Pfades werden, der von einem seiner übergeordneten Knoten ausgeht, oder ein neuer Pfad kann von diesem Knoten selbst ausgehen. Der Schlüssel besteht darin, die Pfadlänge für den linken und rechten Unterbaum rekursiv zu ermitteln und dann das Maximum zurückzugeben. Beim Durchlaufen des Baums müssen einige Fälle berücksichtigt werden, die im Folgenden erläutert werden.
- vorh : speichert den Wert des übergeordneten Knotens. Initialisieren Sie prev mit einem Wert kleiner als der des Wurzelknotens, sodass der Pfad, der bei der Wurzel beginnt, eine Länge von mindestens 1 haben kann.
- nur : Speichert die Pfadlänge, die am übergeordneten Knoten des aktuell besuchten Knotens endet.
Fall 1 : Der Wert des aktuellen Knotens ist vorher +1
Erhöhen Sie in diesem Fall die Pfadlänge um 1, ermitteln Sie dann rekursiv die Pfadlänge für den linken und den rechten Unterbaum und geben Sie dann das Maximum zwischen zwei Längen zurück.
Fall 2 : Der Wert des aktuellen Knotens ist NICHT prev+1
Ein neuer Pfad kann von diesem Knoten aus beginnen. Ermitteln Sie daher rekursiv die Pfadlänge für den linken und rechten Unterbaum. Der Pfad, der am übergeordneten Knoten des aktuellen Knotens endet, kann größer sein als der Pfad, der an diesem Knoten beginnt. Nehmen Sie also das Maximum des Pfads, der an diesem Knoten beginnt und am vorherigen Knoten endet.
Nachfolgend finden Sie die Umsetzung der obigen Idee.
C++// C++ Program to find Maximum Consecutive // Path Length in a Binary Tree #include
// Java Program to find Maximum Consecutive // Path Length in a Binary Tree import java.util.*; class GfG { // To represent a node of a Binary Tree static class Node { Node left right; int val; } // Create a new Node and return its address static Node newNode(int val) { Node temp = new Node(); temp.val = val; temp.left = null; temp.right = null; return temp; } // Returns the maximum consecutive Path Length static int maxPathLenUtil(Node root int prev_val int prev_len) { if (root == null) return prev_len; // Get the value of Current Node // The value of the current node will be // prev Node for its left and right children int cur_val = root.val; // If current node has to be a part of the // consecutive path then it should be 1 greater // than the value of the previous node if (cur_val == prev_val+1) { // a) Find the length of the Left Path // b) Find the length of the Right Path // Return the maximum of Left path and Right path return Math.max(maxPathLenUtil(root.left cur_val prev_len+1) maxPathLenUtil(root.right cur_val prev_len+1)); } // Find length of the maximum path under subtree rooted with this // node (The path may or may not include this node) int newPathLen = Math.max(maxPathLenUtil(root.left cur_val 1) maxPathLenUtil(root.right cur_val 1)); // Take the maximum previous path and path under subtree rooted // with this node. return Math.max(prev_len newPathLen); } // A wrapper over maxPathLenUtil(). static int maxConsecutivePathLength(Node root) { // Return 0 if root is NULL if (root == null) return 0; // Else compute Maximum Consecutive Increasing Path // Length using maxPathLenUtil. return maxPathLenUtil(root root.val-1 0); } //Driver program to test above function public static void main(String[] args) { Node root = newNode(10); root.left = newNode(11); root.right = newNode(9); root.left.left = newNode(13); root.left.right = newNode(12); root.right.left = newNode(13); root.right.right = newNode(8); System.out.println('Maximum Consecutive Increasing Path Length is '+maxConsecutivePathLength(root)); } }
# Python program to find Maximum consecutive # path length in binary tree # A binary tree node class Node: # Constructor to create a new node def __init__(self val): self.val = val self.left = None self.right = None # Returns the maximum consecutive path length def maxPathLenUtil(root prev_val prev_len): if root is None: return prev_len # Get the value of current node # The value of the current node will be # prev node for its left and right children curr_val = root.val # If current node has to be a part of the # consecutive path then it should be 1 greater # than the value of the previous node if curr_val == prev_val +1 : # a) Find the length of the left path # b) Find the length of the right path # Return the maximum of left path and right path return max(maxPathLenUtil(root.left curr_val prev_len+1) maxPathLenUtil(root.right curr_val prev_len+1)) # Find the length of the maximum path under subtree # rooted with this node newPathLen = max(maxPathLenUtil(root.left curr_val 1) maxPathLenUtil(root.right curr_val 1)) # Take the maximum previous path and path under subtree # rooted with this node return max(prev_len newPathLen) # A Wrapper over maxPathLenUtil() def maxConsecutivePathLength(root): # Return 0 if root is None if root is None: return 0 # Else compute maximum consecutive increasing path # length using maxPathLenUtil return maxPathLenUtil(root root.val -1 0) # Driver program to test above function root = Node(10) root.left = Node(11) root.right = Node(9) root.left.left = Node(13) root.left.right = Node(12) root.right.left = Node(13) root.right.right = Node(8) print ('Maximum Consecutive Increasing Path Length is') print (maxConsecutivePathLength(root)) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# Program to find Maximum Consecutive // Path Length in a Binary Tree using System; class GfG { // To represent a node of a Binary Tree class Node { public Node left right; public int val; } // Create a new Node and return its address static Node newNode(int val) { Node temp = new Node(); temp.val = val; temp.left = null; temp.right = null; return temp; } // Returns the maximum consecutive Path Length static int maxPathLenUtil(Node root int prev_val int prev_len) { if (root == null) return prev_len; // Get the value of Current Node // The value of the current node will be // prev Node for its left and right children int cur_val = root.val; // If current node has to be a part of the // consecutive path then it should be 1 greater // than the value of the previous node if (cur_val == prev_val+1) { // a) Find the length of the Left Path // b) Find the length of the Right Path // Return the maximum of Left path and Right path return Math.Max(maxPathLenUtil(root.left cur_val prev_len+1) maxPathLenUtil(root.right cur_val prev_len+1)); } // Find length of the maximum path under subtree rooted with this // node (The path may or may not include this node) int newPathLen = Math.Max(maxPathLenUtil(root.left cur_val 1) maxPathLenUtil(root.right cur_val 1)); // Take the maximum previous path and path under subtree rooted // with this node. return Math.Max(prev_len newPathLen); } // A wrapper over maxPathLenUtil(). static int maxConsecutivePathLength(Node root) { // Return 0 if root is NULL if (root == null) return 0; // Else compute Maximum Consecutive Increasing Path // Length using maxPathLenUtil. return maxPathLenUtil(root root.val - 1 0); } // Driver code public static void Main(String[] args) { Node root = newNode(10); root.left = newNode(11); root.right = newNode(9); root.left.left = newNode(13); root.left.right = newNode(12); root.right.left = newNode(13); root.right.right = newNode(8); Console.WriteLine('Maximum Consecutive' + ' Increasing Path Length is '+ maxConsecutivePathLength(root)); } } // This code has been contributed by 29AjayKumar
<script> // Javascript Program to find Maximum Consecutive // Path Length in a Binary Tree // To represent a node of a Binary Tree class Node { constructor(val) { this.val = val; this.left = this.right = null; } } // Returns the maximum consecutive Path Length function maxPathLenUtil(rootprev_valprev_len) { if (root == null) return prev_len; // Get the value of Current Node // The value of the current node will be // prev Node for its left and right children let cur_val = root.val; // If current node has to be a part of the // consecutive path then it should be 1 greater // than the value of the previous node if (cur_val == prev_val+1) { // a) Find the length of the Left Path // b) Find the length of the Right Path // Return the maximum of Left path and Right path return Math.max(maxPathLenUtil(root.left cur_val prev_len+1) maxPathLenUtil(root.right cur_val prev_len+1)); } // Find length of the maximum path under subtree rooted with this // node (The path may or may not include this node) let newPathLen = Math.max(maxPathLenUtil(root.left cur_val 1) maxPathLenUtil(root.right cur_val 1)); // Take the maximum previous path and path under subtree rooted // with this node. return Math.max(prev_len newPathLen); } // A wrapper over maxPathLenUtil(). function maxConsecutivePathLength(root) { // Return 0 if root is NULL if (root == null) return 0; // Else compute Maximum Consecutive Increasing Path // Length using maxPathLenUtil. return maxPathLenUtil(root root.val-1 0); } // Driver program to test above function let root = new Node(10); root.left = new Node(11); root.right = new Node(9); root.left.left = new Node(13); root.left.right = new Node(12); root.right.left = new Node(13); root.right.right = new Node(8); document.write('Maximum Consecutive Increasing Path Length is '+ maxConsecutivePathLength(root)+'
'); // This code is contributed by rag2127 </script>
Ausgabe
Maximum Consecutive Increasing Path Length is 3
Zeitkomplexität: O(n^2) wobei n die Anzahl der Knoten im gegebenen Binärbaum ist.
Hilfsraum: O(log(n))