Iterative preorder traversal
In last post Iterative inorder traversal , we learned how to do inorder traversal of binary tree without recursion or in iterative way. Today we will learn how to do iterative preorder traversal of binary tree. In preorder traversal, root node is processed before left and right subtrees. For example, preorder traversal of below tree would be [10,5,1,6,14,12,15],
We already know how to implement preorder traversal in recursive way, let’s understand how to implement it in nonrecursive way.
Iterative preorder traversal : Thoughts
If we look at recursive implementation, we see we process the root node as soon as we reach it and then start with left subtree before touching anything on right subtree.
Once left subtree is processed, control goes to first node in right subtree. To emulate this behavior in nonrecursive way, it is best to use a stack. What and when push and pop will happen on the stack?
Start with pushing the root node to stack. Traversal continues till there at least one node onto stack.
Pop the root node from stack,process it and push it’s right and left child on to stack. Why right child before left child? Because we want to process left subtree before right subtree. As at every node, we push it’s children onto stack, entire left subtree of node will be processed before right child is popped from the stack. Algorithm is very simple and is as follows.

 Start with root node and push on to stack s
 While there stack is not empty
 Pop from stack
current
= s.pop() and process the node.  Push
current.right
onto to stack.  Push
current.left
onto to stack.
 Pop from stack
Iterative preorder traversal : example
Let’s take and example and see how it works. Given below tree, do preorder traversal on it without recursion.
Let’s start from root node(10) and push it onto stack. current = node(10).
Here loop starts, which check if there is node onto stack. If yes, it pops that out. s.pop will return node(10), we will print it and push it’s right and left child onto stack. Preorder traversal till now : [10].
Since stack is not empty, pop from it.current= node(5). Print it and push it’s right and left child i.e node(6) and node(1) on stack.
Again, stack is not empty, pop from stack. current = node(1). Print node. There is no right and left child for this node, so we will not push anything on the stack.
Stack is not empty yet, pop again. current= node(6). Print node. Similar to node(1), it also does not have right or left subtree, so nothing gets pushed onto stack.
However, stack is not empty yet. Pop. Current = node(14). Print node, and as there are left and right children, push them onto stack as right child before left child.
Stack is not empty, so pop from stack, current = node(12). Print it, as there are no children of node(12), push nothing to stack.
Pop again from stack as it not empty. current = node(15). Print it. No children, so no need to push anything.
At this point, stack becomes empty and we have traversed all node of tree also.
Iterative preorder traversal : Implementation
#include <stdio.h> #include<stdlib.h> #include<math.h> struct node{ int value; struct node *left; struct node *right; }; typedef struct node Node; #define STACK_SIZE 10 typedef struct stack{ int top; Node *items[STACK_SIZE]; }stack; void push(stack *ms, Node *item){ if(ms>top < STACK_SIZE1){ ms>items[++(ms>top)] = item; } else { printf("Stack is full\n"); } } Node * pop (stack *ms){ if(ms>top > 1 ){ return ms>items[(ms>top)]; } else{ printf("Stack is empty\n"); } } Node * peek(stack ms){ if(ms.top < 0){ printf("Stack empty\n"); return 0; } return ms.items[ms.top]; } int isEmpty(stack ms){ if(ms.top < 0) return 1; else return 0; } void preorderTraversalWithoutRecursion(Node *root){ stack ms; ms.top = 1; if(root == NULL) return ; Node *currentNode = NULL; /* Step 1 : Start with root */ push(&ms,root); while(!isEmpty(ms)){ /* Step 5 : Pop the node */ currentNode = pop(&ms); /* Step 2 : Print the node */ printf("%d ", currentNode>value); /* Step 3: Push right child first */ if(currentNode>right){ push(&ms, currentNode>right); } /* Step 4: Push left child */ if(currentNode>left){ push(&ms, currentNode>left); } } } void preorder (Node *root){ if ( !root ) return; printf("%d ", root>value ); preorder(root>left); preorder(root>right); } Node * createNode(int value){ Node * newNode = (Node *)malloc(sizeof(Node)); newNode>value = value; newNode>right= NULL; newNode>left = NULL; return newNode; } Node * addNode(Node *node, int value){ if(node == NULL){ return createNode(value); } else{ if (node>value > value){ node>left = addNode(node>left, value); } else{ node>right = addNode(node>right, value); } } return node; } /* Driver program for the function written above */ int main(){ Node *root = NULL; //Creating a binary tree root = addNode(root,30); root = addNode(root,20); root = addNode(root,15); root = addNode(root,25); root = addNode(root,40); root = addNode(root,37); root = addNode(root,45); preorder(root); printf("\n"); preorderTraversalWithoutRecursion(root); return 0; }
Complexity of iterative implementation of binary tree is O(n) as we will be visiting each node at least once. Also, there is added space complexity of stack which is O(n).
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