tree-traversal

问题 This question was asked in a recent coding interview. Q : Given a binary tree, write a program to convert it to a doubly linked list. The nodes in the doubly linked list are arranged in a sequence formed by a zig-zag level order traversal My approach i could always do the zig-zag level order traversal of the tree and store it in an array an then make a double linked list. but the question demands for a in-place solution. can anyone help in explaining the recursive approach should be used? 回答1

问题 I wish to print a binary tree in Newick format, showing each node's distance to its parent. At the moment I haven't had an issue with the following code, which uses regular recursion, but a tree too deep may produce a stack overflow. (defn tree->newick [tree] (let [{:keys [id children to-parent]} tree dist (double to-parent)] ; to-parent may be a rational (if children (str "(" (tree->newick (first children)) "," (tree->newick (second children)) "):" dist) (str (name id) ":" dist)))) (def

问题 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 =

问题 void traverse(Node* root) { queue<Node*> q; Node* temp_node= root; while(temp_node) { cout<<temp_node->value<<endl; if(temp_node->left) q.push(temp_node->left); if(temp_node->right) q.push(temp_node->right); if(!q.empty()) { temp_node = q.front(); q.pop(); } else temp_node = NULL; } } The above posted code is my level order traversal code. This code works fine for me but One thing I dont like is I am explicitly initializing temp_node = NULL or I use break. But it does not seem to be a good

问题 I have an array similar to this: Array ( Array ( [ID] => 1 [parentcat_ID] => 0 ), Array ( [ID] => 2 [parentcat_ID] => 0 ), Array ( [ID] => 6 [parentcat_ID] => 1 ), Array ( [ID] => 7 [parentcat_ID] => 1 ), Array ( [ID] => 8 [parentcat_ID] => 6 ), Array ( [ID] => 9 [parentcat_ID] => 1 ), Array ( [ID] => 13 [parentcat_ID] => 7 ), Array ( [ID] => 14 [parentcat_ID] => 8 ) ) But I need a function to recursively put each item into a 'children' array inside the relevant parent array. So it would look

问题 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 what I've written using a stack, but it's not relevant anyway. Here's what I've tried: I attempted to do a pre-order traversal and I got to the left-most node, but I'm stuck there. I don't know how to "recurse" back up without a stack/parent pointer. Any help would be appreciated. (I'm tagging it as Java since that

问题 here's the rough html I get to work with: <li class="par_cat"></li> <li class="sub_cat"></li> <li class="sub_cat"></li> <li class="par_cat"></li> // this is the single element I need to select <li class="sub_cat"></li> <li class="sub_cat"></li> <li class="sub_cat current_sub"></li> // this is where I need to start searching <li class="par_cat"></li> <li class="sub_cat"></li> <li class="par_cat"></li> I need to traverse from the .current_sub , find the closest previous .par_cat and do stuff to

问题 In data structures, I get converting in order and pre-order formula conversions into trees. However, I'm not so good with post-order. For the given formula x y z + a b - c * / - I came up with - / \ * / (divide) / \ / \ x + - c / \ /\ y z a b For the most part, this seems to fit, except the * in the left subtree is the joker in the deck. In post order traversal, the last character is the top node of the tree, everything else branches down. Now I take the / and * operators to mean that they

问题 In a generic tree represented by nodes having pointers to parent, siblings, and firs/last children, as in: class Tnode { def data Tnode parent = null Tnode first_child = null, last_child = null Tnode prev_sibling = null, next_sibling = null Tnode(data=null) { this.data = data } } Is it possible to do an iterative (non-recursive) breadth-first (level-order) traversal without using any additional helper structures such as a queue. So basically: we can use single node references for backtracking

问题 I wanna visit the nodes in an AST from a Java file and extract some information related to these nodes. But, I wanna pass by the AST guided by the lines in the source code file. I know there is information about the lines associated with each node, but the problem is that the default way to access the nodes is through specific visitors. So: 1. to avoid redundant visits to the nodes, 2. do not generates overhead while trying to enumerate all the possible node types (or visitors), and 3. to