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:
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).
Elk knooppunt in de binaire boom kan onderdeel worden van het pad dat begint bij een van de bovenliggende knooppunten, of er kan een nieuw pad beginnen vanaf dit knooppunt zelf. De sleutel is om recursief de padlengte voor de linker en rechter subboom te vinden en vervolgens het maximum terug te geven. Bij het doorkruisen van de boom moet rekening worden gehouden met enkele gevallen, die hieronder worden besproken.
- vorige : slaat de waarde van het bovenliggende knooppunt op. Initialiseer prev met één minder dan de waarde van het hoofdknooppunt, zodat het pad dat begint bij de hoofdmap een lengte van minimaal 1 kan hebben.
- alleen : Slaat de padlengte op die eindigt bij het bovenliggende knooppunt van het momenteel bezochte knooppunt.
Geval 1 : Waarde van huidig knooppunt is vorig +1
In dit geval vergroot u de padlengte met 1 en vindt u vervolgens recursief de padlengte voor de linker en rechter subboom en retourneert u vervolgens het maximum tussen twee lengtes.
Geval 2 : Waarde van huidig knooppunt is NIET vorige+1
Een nieuw pad kan vanaf dit knooppunt beginnen, dus zoek recursief de padlengte voor de linker en rechter subboom. Het pad dat eindigt bij het bovenliggende knooppunt van het huidige knooppunt kan groter zijn dan het pad dat begint bij dit knooppunt. Neem dus het maximum van het pad dat begint bij dit knooppunt en eindigt bij het vorige knooppunt.
Hieronder ziet u de implementatie van bovenstaand 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>
Uitvoer
Maximum Consecutive Increasing Path Length is 3
Tijdcomplexiteit: O(n^2) waarbij n het aantal knooppunten in de gegeven binaire boom is.
Hulpruimte: O(log(n))