This is not homework, this is an interview question.
The catch here is that the algorithm should be constant space.
I'm pretty clueless on how to do this without a stack, I'd post ... |
Okay. I have a binary tree, and this is what I want to do with it:
For each node in original tree:
If it's not a leaf, replace it with a leaf ... |
I have created a class that populates a binary tree with morse code. Where traversing to the left signifies a DOT and traversing to the right signifies a DASH. ... |
I need to traverse a binary tree, skipping the children of any node for which a condition is met.
This implements a tree-guided clustering approach; the leaves of a subtree are considered ... |
i have made an binary tree, and im trying to make an postorder traversal. I have already made an preorder traversal - which i though was what i needed. But after ... |
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Hello, I'm currently trying to work my way through Java and need some advice on using and traversing trees. I've written a basic JTree program, which allows the user to add and delete nodes. Each new node is labelled in a sequential order and not dependent upon where they are added to the tree. Basically, what is the best way to ... |
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I need help! I'm writing code to traverse a tree, find a given character, and then print out it's path labels. It's basically for Huffman coding. I've written the following, but it just goes on and on. Can someone help? Thanks! private static String traverse(HuffNode r, char c) //this method traverses the tree for codewords; code is backwards { String result ... |
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John, Usually the Towers of Hanoi problem is created in such a way as the recursive traversal of the data creates a "binary tree" in code. And usually it is considered in-order. But given a general binary tree one traverses the orders by recrusive moves, such as: for in-order: Traverse( type root ) { if root == null quit Traverse ( ... |
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public void traverse(Node node, List path) { if (node != null) { // if we haven't fallen off the tree list.add(node); if (node.left == null && node.right == null) // it's a leaf swapPayloads(path); else { traverse(node.right, path); // right first, left last traverse(node.left, path); } list.remove(list.size()-1); // remove last node from the path } } |
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Hi I've written the following code for converting Postfix to Infix,it reads a string then reads it letter by letter ,a stack and a Tree is used,and I want to traverse the last object that has poped from the stack,in the following code I used this line: System.out.println((Character)(((TreeNode)stack.pop()).data)); but it only prints the root of the tree,I tried to traverse it ... |
no its a study question for a test....supposedly there is going to be a problem similar and i am just having trouble understanding it. I know the answer is for the entire thing is A / \ C B / \ \ F E D / \ \ I H G im just not understanding the steps very well |
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Hi, I need to develop a code which traverse multi nodes of tree .can i get the code for it.please help me in this regard. i am giving example below. News -| 1.General news 2.Sports news 3.Film News - - -| 1. Holliwood news 2.Bolliwood news 3.Tolliwood news - - | New film by Raja mouli is going to coem on ... |
The level will add hence for your example it will give me level 0 - 2 level 1 - 1 level 2 - 3 But for node 3 it shouldn't be level 2, it should be level 1. May i know what wrong with my code? Or can you provide any explaination or eaxmple? |
I'm importing a separate class to construct my tree. Sorry if that wasn't clear. I'm trying to implement the method I showed before but it doesn't seem like there is a way to get a hold of the root node. This is my first project with binary search trees and I've run out of ideas. Any suggestions... |
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Hi all, I'm trying to traverse a DefaultMutableNode tree using breadthFirstEnumeration. However, while traversing I need to remove some branches and continue with the Enumeration. Do you know any way to preserve the Enumeration after removing a branch??? For example: for(Enumeration en = trial.breadthFirstEnumeration(); en.hasMoreElements() { DefaultMutableTreeNode currentNode = (DefaultMutableTreeNode)en.nextElement(); if((String)currentNode.getUserObject() == "2") { ((DefaultMutableTreeNode)(currentNode)).removeAllChildren(); } System.out.println((String)currentNode.getUserObject()); } This code traverses ... |