binary-tree

It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to

I know that, BST does not allow duplicates. For example, if I have a word "RABSAB". The Binary search tree for the above string is: R /\ A S \ B What if we wanted to include the duplicates in the tree. How the tree gonna change? I was asked this question in an interview. They asked me to draw: a binary tree an unbalanced Binary Search Tree a binary search tree without duplicates a binary search tree with duplicates Any Help is appreciated! PS: Help me by drawing the related trees Sazzadur Rahaman Rule to insert in a binary Search tree without duplicate is: Go left if element is less than root

It seems to me that the only advantage of heap over binary tree is to find the smallest item in the heap in complexity of O(1) instead of O(log(2)n) in binary tree. When implementing priority queue you need to delete the smallest item each from the data structre. deleting the smallest item from a tree and both heap done in complexity of O(log(2)n). Althogh deleting item from a tree may be more complex. Deleting item with no childrens acctually very simple. My question is why use heap instead of binary tree(which is simpler in this case) when implementing priority queue? Worst case complexity

This is not a homework. Just an interesting task :) Given a complete binary search three represensted by array. Sort the array in O(n) using constant memory. Example: Tree: 8 / \ 4 12 /\ / \ 2 6 10 14 /\ /\ /\ /\ 1 3 5 7 9 11 13 15 Array: 8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15 Output: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 It is possible, people calling it homework probably haven't tried solving it yet. We use the following as a sub-routine: Given an array a1 a2 ... an b1 b2 .. bn, convert in O(n) time and O(1) space to b1 a1 b2 a2 ... bn an A solution for that can be

Is there a java library that has Binary Tree that I can use? I am not looking forward to test and implement my own. The Java standard API only contains libraries that are universally useful and non-trivial to implement. A basic tree is trivial to implement: class BinaryTree { BinaryTree left; BinaryTree right; Object value; } Non-trivial trees are not universally useful: either they are needed as a part of the application data model, which is better modeled using domain specific classes (component has-a list of sub-components), or they are used as a part of a specific algorithm. Algorithms

I understand that the degree of a node is the number of children it has. However, how do we define the degree of a tree? Basically The degree of the tree is the total number of it's children i-e the total number nodes that originate from it.The leaf of the tree doesnot have any child so its degree is zero. The degree of a node is the number of partitions in the subtree which has that node as the root. Nodes with degree=0 are called leaves. In general a graph has a minimum degree and a maximum degree, that is just the minimum respectivly the maximum degree of all nodes in the graph. If a graph

I'm trying to insert a Binary Node. My code is convoluted and there's no hope of rescuing it so I plan to rewrite it (basically I didn't account for backtracking and didn't think about the algorithm all that closely). I'm trying to insert a Binary Node using in order traversal, but I don't understand how I'm supposed to backtrack. D / \ B E / \ / \ A C F How would I search through the left subtree of root D and then go back and search through the right one? It might be a stupid question, but I'm stumped. The best I can come up with is something like this: if (!root.hasLeftChild) { root = root