inorder

问题 When I am trying to print level Order of BST, this question prompted me. Here is a Pre-Order Sequence: 4, 1, 2, 3, 5, 6, 7, 8 In_order Sequence : 1, 2, 3, 4, 5, 6, 7, 8 A level order sequence for a BST with above pre_order and In_order is [4, 2, 6, 1, 3, 5, 7, 8] However, for the same Pre-order an In-order sequence this level order sequence seems possible. [4, 1, 5, 2, 6, 3, 7, 8] . I don't know how. I am trying to figure this out. I am unable to construct BST in paper (drawing) that

问题 I am trying to perform an inorder traversal of a tree. The code itself feels right, except it is not working properly. I have a feeling it has to either do with the if condition, how append works in python, or something perhaps with return. This works correctly if I use print instead of return, I think, but I want to be able to use return and still get the correct answer. For example, for the tree [1,None,2,3], my code returns [1] which is clearly incorrect. Additionally is it possible to

问题 I am trying to iterate through a tree inorder in Lisp. So far, I managed building the Postorder traversal but inorder gives me headaches.. The tree's format is this: A / \ B C (A 2 B 0 C 2 D 0 E 0) / \ D E (defun traverseTreeMain (l) (traverseTree l nil) ) (defun traverseTree (l lc) (cond ((and (null l) (null lc)) nil) ((null l) (append nil (traverseTree lc nil))) ((=(cadr l) 0) (append (list (car l)) (traverseTree (cddr l) lc) )) ((= (cadr l) 1) (append nil (traverseTree (cddr l) (append (

问题 Can someone please explain to me how recursion works in in Order traversal. here's my inOrder() method. public void inOrder(BinaryNode p){ if(p.left!=null){ inOrder(p.left); } visit(p); if(p.right!=null){ inOrder(p.right); } } public void visit(BinaryNode p){ System.out.println(p.element); } BinaryTree t=new BinaryTree(); t.insert(5); t.insert(t.root,4); t.insert(t.root,6); t.insert(t.root,60); t.insert(t.root,25); t.insert(t.root,10); t.inOrder(t.root); The method inOrder() prints the

问题 Can someone please explain to me how recursion works in in Order traversal. here's my inOrder() method. public void inOrder(BinaryNode p){ if(p.left!=null){ inOrder(p.left); } visit(p); if(p.right!=null){ inOrder(p.right); } } public void visit(BinaryNode p){ System.out.println(p.element); } BinaryTree t=new BinaryTree(); t.insert(5); t.insert(t.root,4); t.insert(t.root,6); t.insert(t.root,60); t.insert(t.root,25); t.insert(t.root,10); t.inOrder(t.root); The method inOrder() prints the

问题 The problem I am tackle with is to find the first occurrence node in its inorder traversal in a BST. The code I have is given below def Inorder_search_recursive(node,key): if not node: return None InOrder_search_recursive(node.lChild) if node.value==key: return node InOrder_search_recursive(node.rChild) This code always return None, what's wrong with it. I think I've return node when I find a node with value k. Why cannot python pass this node???Thanks in advance 回答1: When you call yourself

问题 Recently, my questions were marked duplicate, like this , even if they weren't. So, let me start with following and then I'll explain my question. Why this question is not a duplicate? I'm not asking how to create a binary tree when inorder & preorder traversal are given. I'm asking for the proof, that inorder+preorder traversal define a unique binary tree. Now, to original question . I went to an interview and interviewer asked me this question. I was stuck and couldn't proceed. :| Question:

问题 I have a binary search tree where i have to implement a method called int valueAtPosition(int x) The problem is, that i need the position in an in order traversal. To find the in order traversal i have this the following code, but i don't know how i count the recursive calls, to get the right position. public void inOrderTraverseTree(Node root){ if(root != null){ inOrderTraverseTree(root.leftChild); System.out.println(root); inOrderTraverseTree(root.rightChild); } } 回答1: You can also use a

问题 EDIT This has been resolved by using StringBuilder as suggested in this thread. Thank you :D Hello, I have a tree and am trying to return a String of the content in order. I can currently print out the tree with something like this: public void inOrder() { if (left != null) left.inOrder(); System.out.print(content + " "); if (right != null) right.inOrder(); } But what I want to do is return the String (rather than print out each nodes content while recursing) and I can't work out how to do it

问题 Is there a data structure that can be traversed in both order of insertion and order of magnitude in O(n) with at most O(log(n)) insertion and deletion? In other words given elements 5, 1, 4, 3, 2 (inserted in this order), it can be traversed either as 1,2,3,4,5 or as 5,1,4,3,2 in O(n) time. Of course I could use an array and simply sort it before traversing, but this would require an O(n*log(n)) pre-traversal step. Also, I could use a multi-linked list to achieve O(n) traversal, but in this