Each Answer to this Q is separated by one/two green lines.

I am able to understand preorder traversal without using recursion, but I’m having a hard time with inorder traversal. I just don’t seem to get it, perhaps, because I haven’t understood the inner working of recursion.

This is what I’ve tried so far:

```
def traverseInorder(node):
lifo = Lifo()
lifo.push(node)
while True:
if node is None:
break
if node.left is not None:
lifo.push(node.left)
node = node.left
continue
prev = node
while True:
if node is None:
break
print node.value
prev = node
node = lifo.pop()
node = prev
if node.right is not None:
lifo.push(node.right)
node = node.right
else:
break
```

The inner while-loop just doesn’t feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn’t feel right. Where am I going wrong?

I haven’t tried postorder traversal, but I guess it’s similar and I will face the same conceptual blockage there, too.

Thanks for your time!

P.S.: Definitions of `Lifo`

and `Node`

:

```
class Node:
def __init__(self, value, left=None, right=None):
self.value = value
self.left = left
self.right = right
class Lifo:
def __init__(self):
self.lifo = ()
def push(self, data):
self.lifo = (data, self.lifo)
def pop(self):
if len(self.lifo) == 0:
return None
ret, self.lifo = self.lifo
return ret
```

Start with the recursive algorithm (pseudocode) :

```
traverse(node):
if node != None do:
traverse(node.left)
print node.value
traverse(node.right)
endif
```

This is a clear case of tail recursion, so you can easily turn it into a while-loop.

```
traverse(node):
while node != None do:
traverse(node.left)
print node.value
node = node.right
endwhile
```

You’re left with a recursive call. What the recursive call does is push a new context on the stack, run the code from the beginning, then retrieve the context and keep doing what it was doing. So, you create a stack for storage, and a loop that determines, on every iteration, whether we’re in a “first run” situation (non-null node) or a “returning” situation (null node, non-empty stack) and runs the appropriate code:

```
traverse(node):
stack = []
while !empty(stack) || node != None do:
if node != None do: // this is a normal call, recurse
push(stack,node)
node = node.left
else // we are now returning: pop and print the current node
node = pop(stack)
print node.value
node = node.right
endif
endwhile
```

The hard thing to grasp is the “return” part: you have to determine, in your loop, whether the code you’re running is in the “entering the function” situation or in the “returning from a call” situation, and you will have an `if/else`

chain with as many cases as you have non-terminal recursions in your code.

In this specific situation, we’re using the node to keep information about the situation. Another way would be to store that in the stack itself (just like a computer does for recursion). With that technique, the code is less optimal, but easier to follow

```
traverse(node):
// entry:
if node == NULL do return
traverse(node.left)
// after-left-traversal:
print node.value
traverse(node.right)
traverse(node):
stack = [node,'entry']
while !empty(stack) do:
[node,state] = pop(stack)
switch state:
case 'entry':
if node == None do: break; // return
push(stack,[node,'after-left-traversal']) // store return address
push(stack,[node.left,'entry']) // recursive call
break;
case 'after-left-traversal':
print node.value;
// tail call : no return address
push(stack,[node.right,'entry']) // recursive call
end
endwhile
```

Here is a simple in-order non-recursive c++ code ..

```
void inorder (node *n)
{
stack s;
while(n){
s.push(n);
n=n->left;
}
while(!s.empty()){
node *t=s.pop();
cout<<t->data;
t=t->right;
while(t){
s.push(t);
t = t->left;
}
}
}
```

def print_tree_in(root): stack = [] current = root while True: while current is not None: stack.append(current) current = current.getLeft(); if not stack: return current = stack.pop() print current.getValue() while current.getRight is None and stack: current = stack.pop() print current.getValue() current = current.getRight();

```
def traverseInorder(node):
lifo = Lifo()
while node is not None:
if node.left is not None:
lifo.push(node)
node = node.left
continue
print node.value
if node.right is not None:
node = node.right
continue
node = lifo.Pop()
if node is not None :
print node.value
node = node.right
```

PS: I don’t know Python so there may be a few syntax issues.

Here is a sample of in order traversal using stack in c# (.net):

(for post order iterative you may refer to: Post order traversal of binary tree without recursion)

```
public string InOrderIterative()
{
List<int> nodes = new List<int>();
if (null != this._root)
{
Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
var iterativeNode = this._root;
while(iterativeNode != null)
{
stack.Push(iterativeNode);
iterativeNode = iterativeNode.Left;
}
while(stack.Count > 0)
{
iterativeNode = stack.Pop();
nodes.Add(iterativeNode.Element);
if(iterativeNode.Right != null)
{
stack.Push(iterativeNode.Right);
iterativeNode = iterativeNode.Right.Left;
while(iterativeNode != null)
{
stack.Push(iterativeNode);
iterativeNode = iterativeNode.Left;
}
}
}
}
return this.ListToString(nodes);
}
```

Here is a sample with visited flag:

```
public string InorderIterative_VisitedFlag()
{
List<int> nodes = new List<int>();
if (null != this._root)
{
Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
BinaryTreeNode iterativeNode = null;
stack.Push(this._root);
while(stack.Count > 0)
{
iterativeNode = stack.Pop();
if(iterativeNode.visted)
{
iterativeNode.visted = false;
nodes.Add(iterativeNode.Element);
}
else
{
iterativeNode.visted = true;
if(iterativeNode.Right != null)
{
stack.Push(iterativeNode.Right);
}
stack.Push(iterativeNode);
if (iterativeNode.Left != null)
{
stack.Push(iterativeNode.Left);
}
}
}
}
return this.ListToString(nodes);
}
```

the definitions of the binarytreenode, listtostring utility:

```
string ListToString(List<int> list)
{
string s = string.Join(", ", list);
return s;
}
class BinaryTreeNode
{
public int Element;
public BinaryTreeNode Left;
public BinaryTreeNode Right;
}
```

Simple iterative inorder traversal without recursion

```
'''iterative inorder traversal, O(n) time & O(n) space '''
class Node:
def __init__(self, value, left = None, right = None):
self.value = value
self.left = left
self.right = right
def inorder_iter(root):
stack = [root]
current = root
while len(stack) > 0:
if current:
while current.left:
stack.append(current.left)
current = current.left
popped_node = stack.pop()
current = None
if popped_node:
print popped_node.value
current = popped_node.right
stack.append(current)
a = Node('a')
b = Node('b')
c = Node('c')
d = Node('d')
b.right = d
a.left = b
a.right = c
inorder_iter(a)
```

State can be remembered implicitly,

```
traverse(node) {
if(!node) return;
push(stack, node);
while (!empty(stack)) {
/*Remember the left nodes in stack*/
while (node->left) {
push(stack, node->left);
node = node->left;
}
/*Process the node*/
printf("%d", node->data);
/*Do the tail recursion*/
if(node->right) {
node = node->right
} else {
node = pop(stack); /*New Node will be from previous*/
}
}
}
```

@Victor, I have some suggestion on your implementation trying to push the state into the stack. I don’t see it is necessary. Because every element you take from the stack is already left traversed. so instead of store the information into the stack, all we need is a flag to indicate if the next node to be processed is from that stack or not. Following is my implementation which works fine:

```
def intraverse(node):
stack = []
leftChecked = False
while node != None:
if not leftChecked and node.left != None:
stack.append(node)
node = node.left
else:
print node.data
if node.right != None:
node = node.right
leftChecked = False
elif len(stack)>0:
node = stack.pop()
leftChecked = True
else:
node = None
```

Little Optimization of answer by @Emadpres

```
def in_order_search(node):
stack = Stack()
current = node
while True:
while current is not None:
stack.push(current)
current = current.l_child
if stack.size() == 0:
break
current = stack.pop()
print(current.data)
current = current.r_child
```

This may be helpful (Java implementation)

```
public void inorderDisplay(Node root) {
Node current = root;
LinkedList<Node> stack = new LinkedList<>();
while (true) {
if (current != null) {
stack.push(current);
current = current.left;
} else if (!stack.isEmpty()) {
current = stack.poll();
System.out.print(current.data + " ");
current = current.right;
} else {
break;
}
}
}
```

```
class Tree:
def __init__(self, value):
self.left = None
self.right = None
self.value = value
def insert(self,root,node):
if root is None:
root = node
else:
if root.value < node.value:
if root.right is None:
root.right = node
else:
self.insert(root.right, node)
else:
if root.left is None:
root.left = node
else:
self.insert(root.left, node)
def inorder(self,tree):
if tree.left != None:
self.inorder(tree.left)
print "value:",tree.value
if tree.right !=None:
self.inorder(tree.right)
def inorderwithoutRecursion(self,tree):
holdRoot=tree
temp=holdRoot
stack=[]
while temp!=None:
if temp.left!=None:
stack.append(temp)
temp=temp.left
print "node:left",temp.value
else:
if len(stack)>0:
temp=stack.pop();
temp=temp.right
print "node:right",temp.value
```

Here’s an iterative C++ solution as an alternative to what @Emadpres posted:

```
void inOrderTraversal(Node *n)
{
stack<Node *> s;
s.push(n);
while (!s.empty()) {
if (n) {
n = n->left;
} else {
n = s.top(); s.pop();
cout << n->data << " ";
n = n->right;
}
if (n) s.push(n);
}
}
```

Here is an iterative Python Code for Inorder Traversal ::

```
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def inOrder(root):
current = root
s = []
done = 0
while(not done):
if current is not None :
s.append(current)
current = current.left
else :
if (len(s)>0):
current = s.pop()
print current.data
current = current.right
else :
done =1
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
inOrder(root)
```

For writing iterative equivalents of these recursive methods, we can first understand how the recursive methods themselves execute over the program’s stack. Assuming that the nodes do not have their parent pointer, we need to manage our own “stack” for the iterative variants.

One way to start is to see the recursive method and mark the locations where a call would “resume” (fresh initial call, or after a recursive call returns). Below these are marked as “RP 0”, “RP 1” etc (“Resume Point”). Take example of inorder traversal. (I will present in C language, but same methodology applies to any general language):

```
void in(node *x)
{
/* RP 0 */
if(x->lc) in(x->lc);
/* RP 1 */
process(x);
if(x->rc) in(x->rc);
/* RP 2 */
}
```

Its iterative variant:

```
void in_i(node *root)
{
node *stack[1000];
int top;
char pushed;
stack[0] = root;
top = 0;
pushed = 1;
while(top >= 0)
{
node *curr = stack[top];
if(pushed)
{
/* type (x: 0) */
if(curr->lc)
{
stack[++top] = curr->lc;
continue;
}
}
/* type (x: 1) */
pushed = 0;
process(curr);
top--;
if(curr->rc)
{
stack[++top] = curr->rc;
pushed = 1;
}
}
}
```

The code comments with `(x: 0)`

and `(x: 1)`

correspond to the “RP 0” and “RP 1” resume points in the recursive method. The `pushed`

flag helps us deduce one of these two resume-points. We do not need to handle a node at its “RP 2” stage, so we do not keep such node on stack.

I think part of the problem is the use of the “prev” variable. You shouldn’t have to store the previous node you should be able to maintain the state on the stack (Lifo) itself.

From Wikipedia, the algorithm you are aiming for is:

- Visit the root.
- Traverse the left subtree
- Traverse the right subtree

In pseudo code (disclaimer, I don’t know Python so apologies for the Python/C++ style code below!) your algorithm would be something like:

```
lifo = Lifo();
lifo.push(rootNode);
while(!lifo.empty())
{
node = lifo.pop();
if(node is not None)
{
print node.value;
if(node.right is not None)
{
lifo.push(node.right);
}
if(node.left is not None)
{
lifo.push(node.left);
}
}
}
```

For postorder traversal you simply swap the order you push the left and right subtrees onto the stack.