Using Stack is the obvious way to traverse tree without recursion. Below is an algorithm for traversing binary tree using stack. See this for step wise step execution of the algorithm.

1) Create an empty stack S. 2) Initialize current node as root 3) Push the current node to S and set current = current->left until current is NULL 4) If current is NULL and stack is not empty then a) Pop the top item from stack. b) Print the popped item, set current = popped_item->right c) Go to step 3. 5) If current is NULL and stack is empty then we are done.

Let us consider the below tree for example

1 / \ 2 3 / \ 4 5 Step 1 Creates an empty stack: S = NULL Step 2 sets current as address of root: current -> 1 Step 3 Pushes the current node and set current = current->left until current is NULL current -> 1 push 1: Stack S -> 1 current -> 2 push 2: Stack S -> 2, 1 current -> 4 push 4: Stack S -> 4, 2, 1 current = NULL Step 4 pops from S a) Pop 4: Stack S -> 2, 1 b) print "4" c) current = NULL /*right of 4 */ and go to step 3 Since current is NULL step 3 doesn't do anything. Step 4 pops again. a) Pop 2: Stack S -> 1 b) print "2" c) current -> 5/*right of 2 */ and go to step 3 Step 3 pushes 5 to stack and makes current NULL Stack S -> 5, 1 current = NULL Step 4 pops from S a) Pop 5: Stack S -> 1 b) print "5" c) current = NULL /*right of 5 */ and go to step 3 Since current is NULL step 3 doesn't do anything Step 4 pops again. a) Pop 1: Stack S -> NULL b) print "1" c) current -> 3 /*right of 5 */ Step 3 pushes 3 to stack and makes current NULL Stack S -> 3 current = NULL Step 4 pops from S a) Pop 3: Stack S -> NULL b) print "3" c) current = NULL /*right of 3 */ Traversal is done now as stack S is empty and current is NULL.

## C++

// C++ program to print inorder traversal // using stack. #include<bits/stdc++.h> using namespace std; /* A binary tree Node has data, pointer to left child and a pointer to right child */ struct Node { int data; struct Node* left; struct Node* right; Node (int data) { this->data = data; left = right = NULL; } }; /* Iterative function for inorder tree traversal */ void inOrder(struct Node *root) { stack<Node *> s; Node *curr = root; while (curr != NULL || s.empty() == false) { /* Reach the left most Node of the curr Node */ while (curr != NULL) { /* place pointer to a tree node on the stack before traversing the node's left subtree */ s.push(curr); curr = curr->left; } /* Current must be NULL at this point */ curr = s.top(); s.pop(); cout << curr->data << " "; /* we have visited the node and its left subtree. Now, it's right subtree's turn */ curr = curr->right; } /* end of while */ } /* Driver program to test above functions*/ int main() { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ struct Node *root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); inOrder(root); return 0; }

## C

#include<stdio.h> #include<stdlib.h> #define bool int /* A binary tree tNode has data, pointer to left child and a pointer to right child */ struct tNode { int data; struct tNode* left; struct tNode* right; }; /* Structure of a stack node. Linked List implementation is used for stack. A stack node contains a pointer to tree node and a pointer to next stack node */ struct sNode { struct tNode *t; struct sNode *next; }; /* Stack related functions */ void push(struct sNode** top_ref, struct tNode *t); struct tNode *pop(struct sNode** top_ref); bool isEmpty(struct sNode *top); /* Iterative function for inorder tree traversal */ void inOrder(struct tNode *root) { /* set current to root of binary tree */ struct tNode *current = root; struct sNode *s = NULL; /* Initialize stack s */ bool done = 0; while (!done) { /* Reach the left most tNode of the current tNode */ if(current != NULL) { /* place pointer to a tree node on the stack before traversing the node's left subtree */ push(&s, current); current = current->left; } /* backtrack from the empty subtree and visit the tNode at the top of the stack; however, if the stack is empty, you are done */ else { if (!isEmpty(s)) { current = pop(&s); printf("%d ", current->data); /* we have visited the node and its left subtree. Now, it's right subtree's turn */ current = current->right; } else done = 1; } } /* end of while */ } /* UTILITY FUNCTIONS */ /* Function to push an item to sNode*/ void push(struct sNode** top_ref, struct tNode *t) { /* allocate tNode */ struct sNode* new_tNode = (struct sNode*) malloc(sizeof(struct sNode)); if(new_tNode == NULL) { printf("Stack Overflow \n"); getchar(); exit(0); } /* put in the data */ new_tNode->t = t; /* link the old list off the new tNode */ new_tNode->next = (*top_ref); /* move the head to point to the new tNode */ (*top_ref) = new_tNode; } /* The function returns true if stack is empty, otherwise false */ bool isEmpty(struct sNode *top) { return (top == NULL)? 1 : 0; } /* Function to pop an item from stack*/ struct tNode *pop(struct sNode** top_ref) { struct tNode *res; struct sNode *top; /*If sNode is empty then error */ if(isEmpty(*top_ref)) { printf("Stack Underflow \n"); getchar(); exit(0); } else { top = *top_ref; res = top->t; *top_ref = top->next; free(top); return res; } } /* Helper function that allocates a new tNode with the given data and NULL left and right pointers. */ struct tNode* newtNode(int data) { struct tNode* tNode = (struct tNode*) malloc(sizeof(struct tNode)); tNode->data = data; tNode->left = NULL; tNode->right = NULL; return(tNode); } /* Driver program to test above functions*/ int main() { /* Constructed binary tree is 1 / \ 2 3 / \ 4 5 */ struct tNode *root = newtNode(1); root->left = newtNode(2); root->right = newtNode(3); root->left->left = newtNode(4); root->left->right = newtNode(5); inOrder(root); getchar(); return 0; }

## Java

// non-recursive java program for inorder traversal import java.util.Stack; /* Class containing left and right child of current node and key value*/ class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } /* Class to print the inorder traversal */ class BinaryTree { Node root; void inorder() { if (root == null) return; Stack<Node> s = new Stack<Node>(); Node curr = root; // traverse the tree while (curr != null || s.size() > 0) { /* Reach the left most Node of the curr Node */ while (curr != null) { /* place pointer to a tree node on the stack before traversing the node's left subtree */ s.push(curr); curr = curr.left; } /* Current must be NULL at this point */ curr = s.pop(); System.out.print(curr.data + " "); /* we have visited the node and its left subtree. Now, it's right subtree's turn */ curr = curr.right; } } public static void main(String args[]) { /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.inorder(); } }

## Python

# Python program to do inorder traversal without recursion # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Iterative function for inorder tree traversal def inOrder(root): # Set current to root of binary tree current = root s = [] # initialze stack done = 0 while(not done): # Reach the left most Node of the current Node if current is not None: # Place pointer to a tree node on the stack # before traversing the node's left subtree s.append(current) current = current.left # BackTrack from the empty subtree and visit the Node # at the top of the stack; however, if the stack is # empty you are done else: if(len(s) >0 ): current = s.pop() print current.data, # We have visited the node and its left # subtree. Now, it's right subtree's turn current = current.right else: done = 1 # Driver program to test above function """ Constructed binary tree is 1 / \ 2 3 / \ 4 5 """ root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) inOrder(root) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Time Complexity: O(n)

Output:

4 2 5 1 3

References:

http://web.cs.wpi.edu/~cs2005/common/iterative.inorder

http://neural.cs.nthu.edu.tw/jang/courses/cs2351/slide/animation/Iterative%20Inorder%20Traversal.pps

See this post for another approach of Inorder Tree Traversal without recursion and without stack!

Please write comments if you find any bug in above code/algorithm, or want to share more information about stack based Inorder Tree Traversal.