Iterative post order traversal without keeping visited flag

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Why is it necessary to keep visited flag for iterative post order traversal and not for inorder or pre order iterative traversal.

Is it possible to do post order traversal without keeping visited flag ?

Here's a post-order visit:

{   if not(leaf(t))
    { postordervisit(left(t);
L1: process(t);

It doesn't use any flags. Why do you think a flag is necessary?

Maybe I don't understand the phrase, "iterative post order traversal". By symmetry, if you think that "iterative preorder traversal" doesn't need a flag, I contend you'd have to believe "iterative postorder traversal" is also flag free, and vice-versa.

EDIT: My bad, must be late at night. "Iterative" meaning "implement without recursion". OK, so you implement your own stack of nodes, and you have to implement what amounts to a stack of return addresses. I think the flag is simulating the effect of the return address knowing whether to go to L1 or L2 next. And since you need this for post order, by symmetry you need it for pre-order.

EDIT October 2010: My bad again, the algorithm provided wasn't post-order. Revised.

Do iterative post and in order, Do iterative post and in order traversal without keeping visited flag NULL) { // visit all left nodes of the given node as per postorder //traversal algorithm . push(p ); Post Order traversal of binary tree is: 4, 5, 2, 6, 7, 3, 1 This article is compiled by Aashish Barnwal . Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Postorder traversal iterative version could be implemented without using visited flags, it is just more difficult to implement.

See here for two solutions for iterative postorder traversal without using any visited flags.

Postorder traversal of Binary Tree without , 1) Recursive Postorder Traversal. 2) Postorder traversal We can keep visited flag with every node instead of separate hash table. C++; Java� Do iterative post and in order traversal without keeping visited flag - CUNOMAD August 17, 2009 | Report Duplicate | Flag | PURGE Microsoft Software Engineer / Developer Algorithm Email me when people comment.

Inorder Non-threaded Binary Tree Traversal without Recursion or , Can we do inorder traversal without threads if we have parent pointers Once a node is visited, we need to move to parent also. Initialize a flag: leftdone = false ; 3. If current node is right child, keep moving to ancestors using parent is subtree of another binary tree using preorder traversal : Iterative� Given a binary tree, write iterative and recursive solution to traverse the tree using post-order traversal in C++, Java and Python. Unlike linked lists, one-dimensional arrays and other linear data structures, which are traversed in linear order, trees may be traversed in multiple ways in depth-first order (pre-order, in-order, and post-order) or breadth-first order (level order traversal).

I found a problem in the solution of user 1337c0d3r: it is simply a pre-order in reverse order. My solution uses an "active list" to mark the nodes in the stack.

(You could store the mark flag in the stack. Will post that solution separately. )

void print_postorder(Nodes const& roots)
    typedef std::set<Node const*> ActiveList;
    ActiveList activeNodes;
    vector<Nodes::const_iterator> stack(1, roots.begin());

    while( stack.empty() == false )
        Nodes::const_iterator node = stack.back();
        ActiveList::iterator activeEntry = activeNodes.find( &*node );

        if( activeEntry == activeNodes.end() )
            // This node is now active.
            activeNodes.insert( &*node );
            // Plan to visit each child.
            for( Nodes::const_reverse_iterator rchild = node->children.rbegin();
                 rchild != node->children.rend(); ++rchild )
                Nodes::const_reverse_iterator rchild2 = rchild;
                Nodes::const_iterator child = (++rchild2).base();
            // Post-order visit the node.
            std::size_t depth = activeNodes.size();
            for( std::size_t i = 0; i < 4 * depth; ++i )
                cout << ' ';  // Indent
            cout << node->name << endl;
            // We're finished with this node.
            activeNodes.erase( activeEntry );

// Try this for yourself!  Tree representation:

#include <vector>
#include <set>

struct Node;
typedef std::vector<Node> Nodes;
struct Node
    std::string name;
    Nodes children;

Iterative traversals for Binary Trees, Iterative tree traversals can be coded easily, using a _ visited _ flag. Inorder( Tree *p) { stack < Tree* > S; do { while (p!= Can there be another way in which we can give an in-order traversal without using a stack? a different approach which again needs extra discussion of storing the tree in-memory or otherwise. We have seen different ways of performing postorder traversal on Binary Trees. Post Order Traversal. Iterative Postorder Traversal using Two Stacks. Iterative Postorder Traversal using One Stack. Here is another way of performing the postorder traversal on a Binary Tree iteratively using a single stack. Consider the Below Terminologies:

I believe the algorithm show in previous posting for port-order traversal is in as it "processes" the node before visiting the left sub-tree. Postorder Traversal is essentially the same as Reverse Polish Notation in which the operands (leaf nodes or sub-trees) precede the operator (the next higher sub-tree node).

A corrected postorder traversal algorith would be:

{   if null(t) return;    

This would visit the leaf or sub-tree nodes before visiting the root of the sub-tree.

Tree Traversal - Sarva, For an iterative algorithm without stack, we need to modify the tree, and make it right In iterative algorithm, the strategy is to maintain visited flag, otherwise we will never //visited flag is added to node structure for iterative postorder traversal� Given a Binary Tree, write an iterative function to print Preorder traversal of the given binary tree. Refer this for recursive preorder traversal of Binary Tree. To convert an inherently recursive procedures to iterative, we need an explicit stack. Following is a simple stack based iterative process to print Preorder traversal.

Post order traversal of binary tree without recursion, While the stack is not empty, we keep pushing the left child of the node from top of stack. If the left child does not exist, Iterative Postorder Traversal, Output: 1 4 7 6 3 13 14 10 8. Here's the version with one stack and without a visited flag: See this post for another approach of Inorder Tree Traversal without recursion and without stack! Please write comments if you find any bug in above code/algorithm, or want to share more information about stack based Inorder Tree Traversal.

Postorder traversal of Binary Tree without recursion , Our task is to print the postorder traversal of the binary tree without using We will add a visited flag to every node it the tree itself to reduce the� Output: 1 4 7 6 3 13 14 10 8 Time complexity of above solution is O(n 2) in worst case we move pointer back to head after visiting every node. Alternate solution using unordered_map in which we do not have to move pointer back to head, so time complexity is O(n).

[PDF] Tree Iterators, We have already discussed various tree traversals: pre-order, in-order, post- order. ▫ So, it is Warning: in providing an iterator to client code the BST may not always the recursive function and these build up on the stack as the calls go deeper also for each node a Boolean flag that indicates whether we have visited its. In inorder traversal, the last node visited before the current node is the parent node. If we keep these nodes on some structure, where we can refer them back, things will be easy. As we refer the most recent node added to structure first (when finding parent of the node, we have to just look at the last visited node), Stack is a great

  • Are pointers or references to parent nodes available, or do you need to use an explicit stack?
  • I presume by iterative he means that you implement postordervisit without any recursive calls.
  • I suspect I answered it a way that still leaves it as a homework question :-}
  • Either the Wikipedia ( lies, or your traversal is not post order.
  • @Maciej: Ooops! Revised.