Binary Search Tree In-Order Traversal Iterative Solution

April 20, 2010

Given a binary search tree, print the elements in-order iteratively without using recursion.

Note:
Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, coding a post-order iterative version is a challenge. See my post: Binary Tree Post-Order Traversal Iterative Solution for more details and an in-depth analysis of the problem.

We know the elements can be printed in-order easily using recursion, as follow:

Excessive recursive function calls may cause memory to run out of stack space and extra overhead. Since the depth of a balanced binary search tree is about lg(n), you might not worry about running out of stack space, even when you have a million of elements. But what if the tree is not balanced? Then you are asking for trouble, because in the worst case the height of the tree may go up to n. If that is the case, stack space will eventually run out and your program will crash.

To solve this issue, we need to develop an iterative solution. The idea is easy, we need a stack to store previous nodes, and a visited flag for each node is needed to record if the node has been visited before. When a node is traversed for the second time, its value will be printed. After its value is printed, we push its right child and continue from there.

Alternative Solution:
The above solution requires modification to the original BST data structure (ie, adding a visited flag). The other solution which doesn’t modify the original structure is with the help of a current pointer in addition of a stack.

First, the current pointer is initialized to the root. Keep traversing to its left child while pushing visited nodes onto the stack. When you reach a NULL node (ie, you’ve reached a leaf node), you would pop off an element from the stack and set it to current. Now is the time to print current’s value. Then, current is set to its right child and repeat the process again. When the stack is empty, this means you’re done printing.

We can even do better by refactoring the above code. The refactoring relies on one important observation:

The last traversed node must not have a right child.

Why this is true? To prove this, we assume the opposite, that is: the last traversed node has a right child. This is certainly incorrect, as in-order traversal would have to traverse its right child next before the traversal is done. Since this is incorrect, the last traversed node must not have a right child by contradiction.

Below is the refactored code:

A threaded tree, with the special threading links shown by dashed arrows. A threaded binary tree makes it possible to traverse the values in the binary tree via a linear traversal that is more rapid than a recursive in-order traversal.

Further Thoughts:
The above solutions require the help of a stack to do in-order traversal. Is it possible to do in-order traversal without a stack?

The answer is yes, it’s possible. There’s 2 possible ways that I know of:

1. By adding a parent pointer to the data structure, this allows us to return to a node’s parent (Credits to my friend who provided this solution to me). To determine when to print a node’s value, we would have to determine when it’s returned from. If it’s returned from its left child, then you would print its value then traverse to its right child, on the other hand if it’s returned from its right child, you would traverse up one level to its parent.

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Binary Search Tree In-Order Traversal Iterative Solution, 4.6 out of 5 based on 21 ratings

36 responses to Binary Search Tree In-Order Traversal Iterative Solution

1. Hi,

Inorder can be done without visited flag as well.

void iterative_inorder(Node * node) {

Stack s; // Initially empty

while(1) {
while(node) {
s.push(node);
node = node->left;
}

if(s.isempty()) break;
else {
Node * temp = s.pop();
printf("%d ", temp->value);
node = temp->right;
}
}
}

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• I think this solution is the same as the solution posted by the author using current pointer.

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• I think the outmost while-loop (ie, while(1) is not necessary. Instead, it might cause looping problem when the inner loop is terminated by the break statement. My understand is break statement can only exit the immediate loop in case of nested loops. Once the inner while loop is terminated, there’s no way to terminate the outer loop. I didn’t try this on computer though so I might be wrong.

Correct me if I am wrong. Thanks.

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• Sorry, I was wrong. I don’t think we cannot simply get rid of the outer loop otherwise only a part of the tree is traversed. I change the code as below:

Let me know if you see any problems.

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• oops, it looks like the “submit comment” function eats up all”"s (less than T greater than) in my post.

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2. Hi Prateek,

Yes, I did realize that there is a solution without using a flag, but it's different than yours.

I believe your solution is to traverse the tree via left child down until it reaches the left leave, while pushing the nodes to the stack. Once it reaches the leave, pop off the stack and print its value. Then, point the current node to its right child and repeat again. (That is, treat its right child as another subtree and continue from there).

Thanks for sharing your solution with us, it's appreciated

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3. Hi,

Yes my solution is this only. Do you have a better solution without using flags? If yes, then please post that. I could think of just this approach only.

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4. Prateek,

I've updated this post with my solution. I don't think my solution is better, I think in general the idea is the same as yours. Congrats on solving this problem without using flags.

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5. Hi 1337c0d3r,

Thanks for the blog. Your site has been a great help in preparation.

Prateek

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6. Hi 1337c0d3r,

Thanks for your site. I got into Amazon
And your site was of a great help in achieving that.

Prateek

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7. Hi Prateek,

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8. This works fine for in-order traversal (except that you are pushing a node to the stack unnecessarily – even though that makes your code simpler).

However, I find it quite challenging to apply this construct to a post-order traversal. Have you thought of a common approach that can be applied to all three depth-first traversals in iterative way? I was a very interesting exercise for me.

About traversing the tree in-order without using a stack, it is possible, with partially threaded tree – i.e., only using the right child node pointers to thread the tree. You can even restore the tree to the original shape.

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9. This is the solution w/o recursion and iteration:

void InorderTraverseNoRecursiveIteration(BinaryTree * root)
{
if (!root)
{
return;
}
BinaryTree * prev = NULL;
BinaryTree * cur = root;
while (cur)
{
if (!prev || prev->left == cur || prev->right == cur)
{
if (cur->left)
{
cur = cur->left;
}
else if(cur->right)
{
cout <data <right;
}
else
{
cur = prev;
}
}
else if (cur->left == prev)
{
cout <data <right)
{
cur = cur->right;
}
else
{
cur = cur->parent;
}
}
else if (cur->right == prev)
{
cur = cur->parent;
}
prev = cur;
}
}

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• void InorderTraverseNoRecursiveIteration(BinaryTree * root)
{
if (!root)
{
return;
}
BinaryTree * prev = NULL;
BinaryTree * cur = root;
while (cur)
{
if (!prev || prev->left == cur || prev->right == cur)
{
if (cur->left)
{
cur = cur->left;
}
else if(cur->right)
{
cout <data <right;
}
else
{
cur = prev;
}
}
else if (cur->left == prev)
{
cout <data <right)
{
cur = cur->right;
}
else
{
cur = cur->parent;
}
}
else if (cur->right == prev)
{
cur = cur->parent;
}
prev = cur;
}
}

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10. 1337c0d3,
Why have you not mentioned Morris Traversal for inorder traversal within constant space. Is it not a iterative inorder traversal?

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11. Iterative Inorder with out recursion, the following code works based on the stack size(when it decreases)
public static void inorderIterativeStackSize(TreeNode t) {
if(t != null) {
boolean stackSizeDec;

Stack s = new Stack();
s.push(t);
stackSizeDec = false;

while(s.size() > 0) {
if(stackSizeDec){
if(s.peek().getRight() != null) {
System.out.print(s.peek().getKey()+”, “);
s.push(s.pop().getRight());
stackSizeDec = false;
}
else {
System.out.print(s.pop().getKey()+”, “);
}
}
else {
if(s.peek().getLeft() != null) {
s.push(s.peek().getLeft());
}
else if(s.peek().getRight() != null) {
System.out.print(s.peek().getKey()+”, “);
s.push(s.pop().getRight());
}
else {
stackSizeDec = true;
System.out.print(s.pop().getKey()+”, “);
}
}
}
}

}

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12. Both of the alternative solutions are erroneous.
In the else part the

can only be NULL and you crash.
It should be:

One last note:I don’t get what you mean by the statement

. You don’t seem to use this in the refactoring of the code.

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13. I was wrong in my understanding of the algos. My previous statement is wrong

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14. Dear 1337c0d3r,

1. Go to left extreme, push nodes on the way
2. Start pop-off and process each node till stack exhausts
3. If pop-ed off node has a right child, go to step 1

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17. public void iterativeInorder() {
BSTNode p = root;
Stack<BSTNode> travStack = new Stack<BSTNode>();
while (p != null) {
while(p != null) { // stack the right child (if any)
if (p.right != null) // and the node itself when going
travStack.push(p.right); // to the left;
travStack.push(p);
p = p.left;
}
p = travStack.pop(); // pop a node with no left child
while (!travStack.isEmpty() && p.right == null) { // visit it and all
visit(p); // nodes with no right child;
p = travStack.pop();
}
visit(p); // visit also the first node with
if (!travStack.isEmpty()) // a right child (if any);
p = travStack.pop();
else p = null;
}
}

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18. I think the action to be done in the loop can be divided into 2 parts.
The 1st part is to push left child to the stack; second is to pop the top element and push its right child into the stack. So it should be something like this.

The most difficult part is to figure out the if condition.

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19. VN:F [1.9.22_1171]
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20. VN:F [1.9.22_1171]
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• I believe your code would go in to infinite loop. Consider a simple case:
2
/ \
1 3
\
4
2 gets pushed; top now is 2, it has a left child, then 1 pushed; top is now 1, it has no left child, stack removes 1, 1 gets printed; top now is 2 and it has a left child again!

Please make sure you do run your code before posting it on the Internet.

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• 2
/ \
1 3
\
4
The above tree gets messed up. It should be like this.

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21. Damn it. 4 should be the right child of 3. How my formatting cannot be kept?

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22. A stack solution like in the post order case: enumerate conditions when a node should be printed, and when necessary push a node into the stack for a second time.

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23. Solution with a parent attribute and no stack. This is a plain C solution.

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24. nice work

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25. here is another in order iterative version in python

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26. Your solutions are for a general binary tree. Using the BST property should make the code simpler. Pseudocode below:-

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27. In the solution where structure of tree is changes,
<else {
cout <data <right);
}>
It prints the top element, then pops and then pushes top element’s right child. But as soon as you pop an element, the top is changed. And you will actually skip a node.

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