Givet et binært træ, find længden af den længste sti, som består af noder med på hinanden følgende værdier i stigende rækkefølge. Hver knude betragtes som en sti med længde 1.
Eksempler:
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).
Hver knude i det binære træ kan enten blive en del af stien, som starter fra en af dens overordnede knudepunkter, eller en ny sti kan starte fra denne knude. Nøglen er rekursivt at finde stielængden for venstre og højre undertræ og derefter returnere maksimum. Nogle tilfælde skal overvejes, mens du krydser træet, som diskuteres nedenfor.
- forrige : gemmer værdien af den overordnede node. Initialiser prev med én værdi mindre end rodknudepunktet, så stien, der starter ved rod, kan have en længde på mindst 1.
- kun : Gemmer stielængden, som slutter ved den overordnede for den aktuelt besøgte node.
Tilfælde 1 : Værdien af den aktuelle node er forrige +1
I dette tilfælde øges sti-længden med 1, og find derefter rekursivt sti-længden for venstre og højre undertræ og returner derefter maksimum mellem to længder.
Tilfælde 2 : Værdien af den aktuelle node er IKKE prev+1
En ny sti kan starte fra denne node, så rekursivt find stielængden for venstre og højre undertræ. Stien, der ender ved den nuværende knudes overordnede knude, kan være større end stien, der starter fra denne knude. Så tag det maksimale af stien, som starter fra denne knude, og som slutter ved den forrige knude.
Nedenfor er implementeringen af ovenstående idé.
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>
Produktion
Maximum Consecutive Increasing Path Length is 3
Tidskompleksitet: O(n^2) hvor n er antallet af noder i det givne binære træ.
Hjælpemellemrum: O(log(n))