tree-traversal

Here is the code I wrote to print all paths of a Binary tree from root to leaf: public static void printRootToLeaf(Node1 root, List<Integer> list) { if(root == null) { return; } list.add(root.data); if(root.left == null && root.right == null) { System.out.println(list); return; } printRootToLeaf(root.left, list); printRootToLeaf(root.right, list); } I am calling this method in main like this: public static void main(String[] args) { Node1 root = new Node1(1); Node1 two = new Node1(2); Node1 three = new Node1(3); Node1 four = new Node1(4); Node1 five = new Node1(5); Node1 six = new Node1(6);

I'm having trouble finding a way to build a tree path from a list of related tuples? I only want a list of every path where each node is visited once, aka hamiltonian path. I keep getting close but missing some path. For example, let's say we have this list of connections: connections = [(1, 4), (1, 5), (2, 5), (3, 4), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 4)] desired output: [[1,4,3], [1,4,5,2], [1,5,2], [1,5,4,3], [2,5,1,4,3], [2,5,4,1], [2,5,4,3], [3,4,1,5,2], [3,4,5,1], [3,4,5,2], [4, 3], [4,1,5,2], [4,5,1], [4,5,2], [5, 2], [5,1,4,3], [5,4,1], [5,4,3] ] So each possible path is

When recursively traversing through a directory structure, what is the most efficient algorithm to use if you have more files than directories? I notice that when using depth-first traversal, it seems to take longer when there are a lot of files in a given directory. Does breadth-first traversal work more efficiently in this case? I have no way to profile the two algorithms at the moment so your insights are very much welcome. EDIT: In response to alphazero's comment, I'm using PHP on a Linux machine. Igor Serebryany It makes sense that breadth-first would work better. When you enter your root

Let's say I have a simple binary tree node class, like so: public class BinaryTreeNode { public String identifier = ""; public BinaryTreeNode parent = null; public BinaryTreeNode left = null; public BinaryTreeNode right = null; public BinaryTreeNode(BinaryTreeNode parent, String identifier) { this.parent = parent; //passing null makes this the root node this.identifier = identifier; } public boolean IsRoot() { return parent == null; } } How would I add a method which is able to recursively traverse through any size tree, visiting each and every existing node from left to right, without

I have the following text from an academic course I took a while ago about inorder traversal (they also call it pancaking) of a binary tree (not BST): Inorder tree traversal Draw a line around the outside of the tree. Start to the left of the root, and go around the outside of the tree, to end up to the right of the root. Stay as close to the tree as possible, but do not cross the tree. (Think of the tree — its branches and nodes — as a solid barrier.) The order of the nodes is the order in which this line passes underneath them. If you are unsure as to when you go “underneath” a node,

If the pre-order traversal of a binary search tree is 6, 2, 1, 4, 3, 7, 10, 9, 11, how to get the post-order traversal? You are given the pre-order traversal of the tree, which is constructed by doing: output, traverse left, traverse right. As the post-order traversal comes from a BST, you can deduce the in-order traversal (traverse left, output, traverse right) from the post-order traversal by sorting the numbers. In your example, the in-order traversal is 1, 2, 3, 4, 6, 7, 9, 10, 11. From two traversals we can then construct the original tree. Let's use a simpler example for this: Pre-order:

I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break