Custom Tree Problem - GeeksforGeeks
https://github.com/mission-peace/interview/blob/master/src/com/interview/recursion/PrintArrayInCustomizedFormat.java
void print(char str[][]){
Map<Character,Boolean> visited = new HashMap<Character,Boolean>();
Map<Character,Boolean> alreadyPrinted = new HashMap<Character,Boolean>();
for(int i=0 ; i < str.length; i++){
if(!visited.containsKey(str[i][0]) || !visited.get(str[i][0])){
if(!alreadyPrinted.containsKey(str[i][0]) || !alreadyPrinted.get(str[i][0])){
System.out.println(str[i][0]);
alreadyPrinted.put(str[i][0],true);
}
DFS(str,i,5,visited);
}
}
}
private void DFS(char str[][],int pos,int distance,Map<Character,Boolean> visited){
for(int i=0; i < distance; i++){
System.out.print(" ");
}
System.out.println(str[pos][1]);
char ch = str[pos][1];
visited.put(ch, true);
int i = pos+1;
for(; i < str.length; i++){
if(ch == str[i][0]){
if(i != str.length){
DFS(str,i,distance + 5,visited);
}
}
}
}
Read full article from Custom Tree Problem - GeeksforGeeks
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
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.
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 nodeclass 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 + " |"); } } }void print(char str[][]){
Map<Character,Boolean> visited = new HashMap<Character,Boolean>();
Map<Character,Boolean> alreadyPrinted = new HashMap<Character,Boolean>();
for(int i=0 ; i < str.length; i++){
if(!visited.containsKey(str[i][0]) || !visited.get(str[i][0])){
if(!alreadyPrinted.containsKey(str[i][0]) || !alreadyPrinted.get(str[i][0])){
System.out.println(str[i][0]);
alreadyPrinted.put(str[i][0],true);
}
DFS(str,i,5,visited);
}
}
}
private void DFS(char str[][],int pos,int distance,Map<Character,Boolean> visited){
for(int i=0; i < distance; i++){
System.out.print(" ");
}
System.out.println(str[pos][1]);
char ch = str[pos][1];
visited.put(ch, true);
int i = pos+1;
for(; i < str.length; i++){
if(ch == str[i][0]){
if(i != str.length){
DFS(str,i,distance + 5,visited);
}
}
}
}
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