Custom Tree Problem
You are given a set of links, e.g.
a ---> b b ---> c b ---> d a ---> e
Print the tree that would form when each pair of these links that has the same character as start and end point is joined together. You have to maintain fidelity w.r.t. the height of nodes, i.e. nodes at height n from root should be printed at same row or column. For set of links given above, tree printed should be –
-->a |-->b | |-->c | |-->d |-->e
Note that these links need not form a single tree; they could form, ahem, a forest. Consider the following links
a ---> b a ---> g b ---> c c ---> d d ---> e c ---> f z ---> y y ---> x x ---> w
The output would be following forest.
-->a |-->b | |-->c | | |-->d | | | |-->e | | |-->f |-->g -->z |-->y | |-->x | | |-->w
You can assume that given links can form a tree or forest of trees only, and there are no duplicates among links.
Solution: The idea is to maintain two arrays, one array for tree nodes and other for trees themselves (we call this array forest). An element of the node array contains the TreeNode object that corresponds to respective character. An element of the forest array contains Tree object that corresponds to respective root of tree. It should be obvious that the crucial part is creating the forest here, once it is created, printing it out in required format is straightforward. To create the forest, following procedure is used –
Do following for each input link, 1. If start of link is not present in node array Create TreeNode objects for start character Add entries of start in both arrays. 2. If end of link is not present in node array Create TreeNode objects for start character Add entry of end in node array. 3. If end of link is present in node array. If end of link is present in forest array, then remove it from there. 4. Add an edge (in tree) between start and end nodes of link.
It should be clear that this procedure runs in linear time in number of nodes as well as of links – it makes only one pass over the links. It also requires linear space in terms of alphabet size. Following is Java implementation of above algorithm. In the following implementation characters are assumed to be only lower case characters from ‘a’ to ‘z’.
Implementation:
Java
// Java program to create a custom tree from a given set of links. // The main class that represents tree and has main method public class Tree { private TreeNode root; /* Returns an array of trees from links input. Links are assumed to be Strings of the form "<s> <e>" where <s> and <e> are starting and ending points for the link. The returned array is of size 26 and has non-null values at indexes corresponding to roots of trees in output */ public Tree[] buildFromLinks(String [] links) { // Create two arrays for nodes and forest TreeNode[] nodes = new TreeNode[ 26 ]; Tree[] forest = new Tree[ 26 ]; // Process each link for (String link : links) { // Find the two ends of current link String[] ends = link.split( " " ); int start = ( int ) (ends[ 0 ].charAt( 0 ) - 'a' ); // Start node int end = ( int ) (ends[ 1 ].charAt( 0 ) - 'a' ); // End node // If start of link not seen before, add it two both arrays if (nodes[start] == null ) { nodes[start] = new TreeNode(( char ) (start + 'a' )); // Note that it may be removed later when this character is // last character of a link. For example, let we first see // a--->b, then c--->a. We first add 'a' to array of trees // and when we see link c--->a, we remove it from trees array. forest[start] = new Tree(nodes[start]); } // If end of link is not seen before, add it to the nodes array if (nodes[end] == null ) nodes[end] = new TreeNode(( char ) (end + 'a' )); // If end of link is seen before, remove it from forest if // it exists there. else forest[end] = null ; // Establish Parent-Child Relationship between Start and End nodes[start].addChild(nodes[end], end); } return forest; } // Constructor public Tree(TreeNode root) { this .root = root; } public static void printForest(String[] links) { Tree t = new Tree( new TreeNode( '\0' )); for (Tree t1 : t.buildFromLinks(links)) { if (t1 != null ) { t1.root.printTreeIdented( "" ); System.out.println( "" ); } } } // Driver method to test public static void main(String[] args) { String [] links1 = { "a b" , "b c" , "b d" , "a e" }; System.out.println( "------------ Forest 1 ----------------" ); printForest(links1); String [] links2 = { "a b" , "a g" , "b c" , "c d" , "d e" , "c f" , "z y" , "y x" , "x w" }; System.out.println( "------------ Forest 2 ----------------" ); printForest(links2); } } // Class to represent a tree node class TreeNode { TreeNode []children; char c; // Adds a child 'n' to this node public void addChild(TreeNode n, int index) { this .children[index] = n;} // Constructor public TreeNode( char c) { this .c = c; this .children = new TreeNode[ 26 ];} // Recursive method to print indented tree rooted with this node. public void printTreeIdented(String indent) { System.out.println(indent + "-->" + c); for (TreeNode child : children) { if (child != null ) child.printTreeIdented(indent + " |" ); } } } |
C#
// C# program to create a custom tree // from a given set of links. using System; // The main class that represents tree // and has main method class Tree { public TreeNode root; /* Returns an array of trees from links input. Links are assumed to be Strings of the form "<s> <e>" where <s> and <e> are starting and ending points for the link. The returned array is of size 26 and has non-null values at indexes corresponding to roots of trees in output */ public Tree[] buildFromLinks(String [] links) { // Create two arrays for nodes and forest TreeNode[] nodes = new TreeNode[26]; Tree[] forest = new Tree[26]; // Process each link foreach (String link in links) { char []sep = { ' ' , ' ' }; // Find the two ends of current link String[] ends = link.Split(sep); int start = ( int ) (ends[0][0] - 'a' ); // Start node int end = ( int ) (ends[1][0] - 'a' ); // End node // If start of link not seen before, // add it two both arrays if (nodes[start] == null ) { nodes[start] = new TreeNode(( char ) (start + 'a' )); // Note that it may be removed later when // this character is last character of a link. // For example, let we first see a--->b, // then c--->a. We first add 'a' to array // of trees and when we see link c--->a, // we remove it from trees array. forest[start] = new Tree(nodes[start]); } // If end of link is not seen before, // add it to the nodes array if (nodes[end] == null ) nodes[end] = new TreeNode(( char ) (end + 'a' )); // If end of link is seen before, // remove it from forest if it exists there. else forest[end] = null ; // Establish Parent-Child Relationship // between Start and End nodes[start].addChild(nodes[end], end); } return forest; } // Constructor public Tree(TreeNode root) { this .root = root; } public static void printForest(String[] links) { Tree t = new Tree( new TreeNode( '\0' )); foreach (Tree t1 in t.buildFromLinks(links)) { if (t1 != null ) { t1.root.printTreeIdented( "" ); Console.WriteLine( "" ); } } } // Driver Code public static void Main(String[] args) { String [] links1 = { "a b" , "b c" , "b d" , "a e" }; Console.WriteLine( "------------ Forest 1 ----------------" ); printForest(links1); String [] links2 = { "a b" , "a g" , "b c" , "c d" , "d e" , "c f" , "z y" , "y x" , "x w" }; Console.WriteLine( "------------ Forest 2 ----------------" ); printForest(links2); } } // Class to represent a tree node public class TreeNode { TreeNode []children; char c; // Adds a child 'n' to this node public void addChild(TreeNode n, int index) { this .children[index] = n; } // Constructor public TreeNode( char c) { this .c = c; this .children = new TreeNode[26]; } // Recursive method to print indented tree // rooted with this node. public void printTreeIdented(String indent) { Console.WriteLine(indent + "-->" + c); foreach (TreeNode child in children) { if (child != null ) child.printTreeIdented(indent + " |" ); } } } // This code is contributed by Rajput-Ji |
Python3
#Python code for the above approach # The main class that represents tree # and has main method class TreeNode: def __init__( self , c): self .c = c self .children = [ None ] * 26 def addChild( self , n, index): self .children[index] = n def printTreeIdented( self , indent): print (indent + "-->" + self .c) for child in self .children: if child: child.printTreeIdented(indent + " |" ) class Tree: def __init__( self , root): self .root = root def buildFromLinks( self , links): """ Returns an array of trees from links input. Links are assumed to be strings of the form "<s> <e>" where <s> and <e> are starting and ending points for the link. The returned array is of size 26 and has non-null values at indexes corresponding to roots of trees in output """ nodes = [ None ] * 26 forest = [ None ] * 26 for link in links: start = ord (link[ 0 ]) - ord ( 'a' ) end = ord (link[ 2 ]) - ord ( 'a' ) # If start of link not seen before, add it two both arrays if not nodes[start]: nodes[start] = TreeNode( chr (start + ord ( 'a' ))) forest[start] = Tree(nodes[start]) # If end of link is not seen before, add it to the nodes array if not nodes[end]: nodes[end] = TreeNode( chr (end + ord ( 'a' ))) # If end of link is seen before, remove it from forest if # it exists there. else : forest[end] = None # Establish Parent-Child Relationship between Start and End nodes[start].addChild(nodes[end], end) return forest @staticmethod def printForest(links): t = Tree(TreeNode( '\0' )) for t1 in t.buildFromLinks(links): if t1: t1.root.printTreeIdented("") print () if __name__ = = "__main__" : links1 = [ "a b" , "b c" , "b d" , "a e" ] print ( "------------ Forest 1 ----------------" ) Tree.printForest(links1) links2 = [ "a b" , "a g" , "b c" , "c d" , "d e" , "c f" , "z y" , "y x" , "x w" ] print ( "------------ Forest 2 ----------------" ) Tree.printForest(links2) |
Javascript
// Class to represent a node in the tree class TreeNode { constructor(c) { // The character stored at this node this .c = c; // An array of 26 nodes to represent children of this node this .children = new Array(26).fill( null ); } // Method to add a child node to this node addChild(n, index) { this .children[index] = n; } // Method to print the tree with indentation printTreeIdented(indent) { document.write(indent + "-->" + this .c); for (const child of this .children) { if (child) { child.printTreeIdented(indent + " |" ); } } } } // Class to represent a tree class Tree { constructor(root) { // The root node of this tree this .root = root; } // Method to build trees from the given links buildFromLinks(links) { // An array to store all nodes in the tree let nodes = new Array(26).fill( null ); // An array to store trees in the forest let forest = new Array(26).fill( null ); // Iterating over all links to build the trees for (const link of links) { // Calculating the starting and ending points of the link let start = link.charCodeAt(0) - 'a' .charCodeAt(0); let end = link.charCodeAt(2) - 'a' .charCodeAt(0); // If start of the link is not seen before, add it to both arrays if (!nodes[start]) { nodes[start] = new TreeNode(String.fromCharCode(start + 'a' .charCodeAt(0))); forest[start] = new Tree(nodes[start]); } // If end of the link is not seen before, add it to the nodes array if (!nodes[end]) { nodes[end] = new TreeNode(String.fromCharCode(end + 'a' .charCodeAt(0))); } // If end of the link is seen before, remove it from the forest array if it exists there else { forest[end] = null ; } // Establish Parent-Child Relationship between Start and End nodes[start].addChild(nodes[end], end); } return forest; } // Method to print the forest of trees static printForest(links) { let t = new Tree( new TreeNode( '\0' )); // Building the forest of trees from the given links let forest = t.buildFromLinks(links); // Iterating over all trees in the forest and printing them for (const t1 of forest) { if (t1) { t1.root.printTreeIdented( "" ); document.write( "<br>" ); } } } } let links1 = [ "a b" , "b c" , "b d" , "a e" ]; document.write( "------------ Forest 1 ----------------" + "<br>" ); Tree.printForest(links1); let links2 = [ "a b" , "a g" , "b c" , "c d" , "d e" , "c f" , "z y" , "y x" , "x w" ]; document.write( "------------ Forest 2 ----------------" + "<br>" ); Tree.printForest(links2); |
C++
#include <iostream> #include <vector> using namespace std; // Class to represent a tree node class TreeNode { public : vector<TreeNode*> children; // vector of child nodes char c; // node's value // Constructor TreeNode( char c) { this ->c = c; } // Adds a child 'n' to this node void addChild(TreeNode* n) { this ->children.push_back(n); } // Recursive method to print indented tree rooted with // this node. void printTreeIndented(string indent) { cout << indent << "-->" << c << endl; for ( auto child : children) { if (child != nullptr) { child->printTreeIndented(indent + " |" ); } } } }; // The main class that represents a forest and has main // method class Tree { private : TreeNode* root; public : // Constructor Tree(TreeNode* root) { this ->root = root; } TreeNode* getRoot() { return root; } /* Returns an array of trees from links input. Links are assumed to be Strings of the form "<s> <e>" where <s> and <e> are starting and ending points for the link. The returned array is of size 26 and has non-null values at indexes corresponding to roots of trees in output */ vector<Tree*> buildFromLinks(vector<string> links) { TreeNode* nodes[26] = { nullptr }; // array of tree nodes vector<Tree*> forest( 26, nullptr); // vector of tree roots // Process each link for ( auto link : links) { // Find the two ends of current link char start = link[0]; // Start node char end = link[2]; // End node // If start of link not seen before, add it to // both arrays if (nodes[start - 'a' ] == nullptr) { nodes[start - 'a' ] = new TreeNode(start); forest[start - 'a' ] = new Tree(nodes[start - 'a' ]); } // If end of link is not seen before, add it to // the nodes array if (nodes[end - 'a' ] == nullptr) { nodes[end - 'a' ] = new TreeNode(end); } // If end of link is seen before, remove it from // forest if it exists there. else { forest[end - 'a' ] = nullptr; } // Establish Parent-Child Relationship between // Start and End nodes[start - 'a' ]->addChild(nodes[end - 'a' ]); } return forest; } }; // Function to print a forest void printForest(std::vector<std::string> links) { Tree t( new TreeNode( '\0' )); for (Tree* t1 : t.buildFromLinks(links)) { if (t1 != nullptr) { t1->getRoot()->printTreeIndented( "" ); std::cout << std::endl; } } } int main() { vector<string> links1 = { "a b" , "b c" , "b d" , "a e" }; cout << "------------ Forest 1 ----------------" << endl; printForest(links1); vector<string> links2 = { "a b" , "a g" , "b c" , "c d" , "d e" , "c f" , "z y" , "y x" , "x w" }; cout << "------------ Forest 2 ----------------" << endl; printForest(links2); return 0; } // This code is contributed by divyansh2212 |
------------ Forest 1 ---------------- -->a |-->b | |-->c | |-->d |-->e ------------ Forest 2 ---------------- -->a |-->b | |-->c | | |-->d | | | |-->e | | |-->f |-->g -->z |-->y | |-->x | | |-->w
Time Complexity: O(n) where n is the number of links in the input array.
Auxiliary Space: O(n)
Exercise: In the above implementation, endpoints of input links are assumed to be from set of only 26 characters. Extend the implementation where endpoints are strings of any length.
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