preorder traversal of binary tree in c using iterative solution with node having a parent pointer

any link to source code of this problem

Hello,

The best way to do the iterative version of this naturally recursive code, relies on the usage of a Stack where nodes are pushed and popped according to the “rules” of the traversal being done.

I believe that this code answers your question pretty well

Best regards,

Bruno

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Pasting the code from @kuruma’s link:

```
#include <stdlib.h>
#include <stdio.h>
#include <iostream>
#include <stack>
using namespace std;
/* A binary tree node has data, left child and right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Helper function that allocates a new node with the given data and
NULL left and right pointers.*/
struct node* newNode(int data)
{
struct node* node = new struct node;
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
// An iterative process to print preorder traversal of Binary tree
void iterativePreorder(node *root)
{
// Base Case
if (root == NULL)
return;
// Create an empty stack and push root to it
stack<node *> nodeStack;
nodeStack.push(root);
/* Pop all items one by one. Do following for every popped item
a) print it
b) push its right child
c) push its left child
Note that right child is pushed first so that left is processed first */
while (nodeStack.empty() == false)
{
// Pop the top item from stack and print it
struct node *node = nodeStack.top();
printf ("%d ", node->data);
nodeStack.pop();
// Push right and left children of the popped node to stack
if (node->right)
nodeStack.push(node->right);
if (node->left)
nodeStack.push(node->left);
}
}
// Driver program to test above functions
int main()
{
/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
struct node *root = newNode(10);
root->left = newNode(8);
root->right = newNode(2);
root->left->left = newNode(3);
root->left->right = newNode(5);
root->right->left = newNode(2);
iterativePreorder(root);
return 0;
}
```

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