This is a C Program to perform pre order traversal. Time Complexity: O(n)
Here is source code of the C Program to Perform Preorder Non-Recursive Traversal of a Given Binary Tree. The C program is successfully compiled and run on a Linux system. The program output is also shown below.
#include <stdlib.h>
#include <stdio.h>
#include <stack>
/* 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;
}
Output:
$ gcc PreorderNonRecursive.c $ ./a.out 10 8 3 5 2 2
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