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Depth First Search (DFS) Non-recursive

, 9 Feb 2014
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DFS algorithm to print the element of a Binary Search Tree in order in which just by traversing with a DFS algorithm is possible

Introduction

A friend asked me about an interview question, how to write a non-recursive DFS algorithm for traversing a binary tree. So I decided to share it with you here.

Background 

In the following code, I apply the DFS algorithm to print the element of a Binary Search Tree in order in which just by traversing with a DFS algorithm is possible.

Using the Code

The logic of the algorithm is traverse to left subtrees as much as possible and push them into the stack. When reaching the end, you pop one element from the stack and print it.Then for the node that is popped from the stack again, we should do the same thing (find the left elements) on the right subtree of that node, and we continue until the stack becomes empty.
  
#include <stack> 
struct BTNode
{

    BTNode(int d)
    {
        data = d;
    }

    int data;
    BTNode* right;
    BTNode* left;
}; 

    void DFS(BTNode* root)
    {
        BTNode* tmp = root;
        std::stack<BTNode*> stck;
        stck.push(tmp);

        //Until the stack is not empty, it means there are nodes to traverse
        while(stck.empty() == false)
        {
            while (tmp->left != 0)
            {
                tmp = tmp->left;
                stck.push(tmp);
            }
            while(true)
            {
                tmp = stck.top();
                stck.pop();
                std::cout << tmp->data;
                if(tmp->right != 0)
                {
                    tmp = tmp->right;
                    stck.push(tmp);
                    break;
                }
            }
        }

 void CreateTreeExample(BTNode* root)
    {
        BTNode* b1= new BTNode(1);
        BTNode* b2= new BTNode(2);
        BTNode* b3= new BTNode(3);
        BTNode* b4= new BTNode(4);
        BTNode* b5= new BTNode(5);
        BTNode* b6= new BTNode(6);
        BTNode* b7= new BTNode(7);
        BTNode* b8= new BTNode(8);
        BTNode* b9= new BTNode(9);

        b2->left = b1;
        b2->right = b4;
        b6->left = b2;
        b4->left = b3;
        b4->right = b5;
        b7->left = b6;
        b7->right = b8;
        b8->right = b9;
        root = b7;
    } 

To test the algorithm, make a simple binary tree by calling the method:
//             7
//            /  \
//           6    8
//          /      \ 
//        2         9
//       /  \
//      1    4
//          /  \
//         3    5

BTNode* root;
CreateTreeExample(root) ;  

and call the DFS method. The output for the example binary tree is:

1
2
3
4
5
6
7
8
9

Points of Interest

FYI, you can use the same concept to solve the question:

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.

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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About the Author

Shiva Varshovi_
Software Developer
Canada Canada
No Biography provided

Comments and Discussions

 
Questionwhile( true) Pinmembercharlie10027-Jul-14 16:14 
QuestionNext challenge PinmemberYvesDaoust10-Feb-14 0:06 
AnswerRe: Next challenge PinmemberShiva Varshovi_12-Feb-14 15:11 

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