preorder traversal of binary tree

preorder traversal of binary tree in c using iterative solution with node having a parent pointer
any link to source code of this problem


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 :slight_smile:

Best regards,


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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;

// An iterative process to print preorder traversal of Binary tree
void iterativePreorder(node *root)
    // Base Case
    if (root == NULL)

    // Create an empty stack and push root to it
    stack<node *> nodeStack;

    /* 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 =;
       printf ("%d ", node->data);

       // Push right and left children of the popped node to stack
       if (node->right)
       if (node->left)

// Driver program to test above functions
int main()
    /* Constructed binary tree is
          /   \
        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);
  return 0;
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