Iterative Binary Tree Traversal Using Stack (Preorder, Inorder and Postorder)

Introduction to iterative tree traversals

In recursive DFS traversal of binary tree, we have three basic elements to traverse: root node, left subtree, and right subtree. Each traversal process nodes in a different order using recursion, where recursive code is simple and easy to visualize i.e. one function parameter and 3–4 lines of code. So the critical question would be: How do we convert it into iterative code using stack? Let’s think.

Stack is a useful data structure to convert a recursive code into an iterative code. In recursive code, under the hood, compiler uses call stack to convert it into an iterative code. This call stack stores information like local variables, input parameters, the current state of the function call, etc.

Sometimes iterative implementation using stack becomes complex due to many input variables, additional operations, and the complex nature of recursive calls. In some situations, recursion is simple, and we can easily convert it into an iterative code. 

To convert recursive code into an iterative one, we need to mimic what compiler does when we run recursive code! For this purpose, we need to use stack data structure in our code to simulate the execution of recursion.

Let's understand the nature of recursion in DFS traversals

Preorder and inorder traversals are tail-recursive i.e. there are no extra operations after the final recursive call. So implementation using stack is simple and easy to understand. On other side, postorder traversal is non-tail recursive because there is an extra operation after the last recursive calls i.e. we process the root node. So the implementation of postorder using a stack would be a little tricky. But if we follow the basic order of processing nodes, it can be easy to visualize.

To simulate the recursive traversal into an iterative traversal, we need to understand the flow of recursive calls in DFS traversal. We visit each node three times:

  • First time: When recursion visit the node coming from the top. In preorder traversal, we process nodes at this stage.
  • Second time: When recursion backtrack from the left child after visiting all nodes in the left subtree. In inorder traversal, we process nodes at this stage.
  • Third time: When recursion backtrack from the right child after visiting all nodes in the right subtree. In postorder traversal, we process nodes at this stage.

Recursive DFS traversal of binary tree

Let's simulate the iterative implementation of each traversal using stack. We will be using the following binary tree node structure:

class TreeNode
{
    int data
    TreeNode left
    TreeNode right
}

Iterative preorder traversal using stack

Let's revise the preorder traversal: We first process the root node, then traverse the left subtree, and finally, traverse the right subtree. Tracking the flow of recursion will give us better picture.

  • We first process the root. Then go to the left subtree and process the root of the left subtree. Then we continue going left in same way until we reach the leftmost node of tree.
  • If the leftmost node is a node with the right child only: Recursion goes to right child of the current node, processes all nodes in the right subtree using same process, and backtracks to the current node. Now traversal of the subtree rooted at the leftmost node is done, and recursion backtracks to its parent node. A similar process continues for the parent node.
  • If the leftmost node is a leaf node: Then traversal of the leftmost node is done, and recursion backtrack to its parent node. It had already processed the parent node when it was coming from top. So recursion goes to the right child of the parent node, processes all nodes in the right subtree, and backtracks to the parent node. This process continues further in a similar way.

Preorder traversal example

Here is an idea of iterative simulation using stack: When we visit any node first time, we process it, push node into the stack (to process the right subtree of that node later), and go to the left child. If there is no left child, we grab one node from the top of the stack and go to the right child of that node. Now we continue the same process for subtree rooted at the right child of popped node. Let's understand it via clear implementation steps.

Implantation steps of iterative preorder traversal

  • We use a stack treeStack and temporary pointer currNode to track the current node. At start, we initialize currNode with the root node.
  • Now we run a loop till treeStack is not empty or currNode != NULL

Condition 1: If (currNode != NULL)

We process the currNode and push it into treeStack. Now we move to the left child i.e. currNode = currNode->left. This process continues in a loop till we reach the leftmost end of binary tree i.e. currNode == NULL. At this stage, all ancestors of that node will be available on treeStack.

if (currNode != NULL)
{
    process(currNode->data)
    treeStack.push(currNode)
    currNode = currNode->left
}

Condition 2: If (currNode == NULL)

Now we reached the leftmost end of binary tree. So we move to the parent of that node by popping a node from the top of treeStack. At this stage, we need to traverse the right subtree of parent node, so we assign currNode with the right child of popped node and continue the above process. Think!

if (currNode == NULL)
{
    prevNode = treeStack.pop()
    currNode = prevNode->right
}

Pseudocode of the iterative preorder traversal

void iterativePreorder(TreeNode root) 
{
    if(root == NULL)
        return
    Stack<TreeNode> treeStack
    TreeNode currNode = root
    TreeNode prevNode = NULL
    while (treeStack.empty() == false || currNode != NULL) 
    {
        if (currNode != NULL)
        {
            process(currNode->data)
            treeStack.push(currNode)
            currNode = currNode->left
        }
        else
        {
            prevNode = treeStack.pop()
            currNode = prevNode->right
        }
    }
}

Optimized version of the above approach

We can further optimise the above solution by pushing only the right children to the stack. The idea is simple:

  • Inside the loop, If (currNode != NULL), we process the currNode and push its right child to the treeStack before moving to its left child.
  • If (currNode == NULL), we pop a node from treeStack and set the currNode to the popped node.
void iterativePreorder(TreeNode root) 
{
    if(root == NULL)
        return
    Stack<TreeNode> treeStack
    TreeNode currNode = root
    while (treeStack.empty() == false || currNode != NULL) 
    {
        if (currNode != NULL)
        {
            process(currNode->data)
            if (currNode->right != NULL)
                treeStack.push(currNode->right)
            currNode = currNode->left
        }
        else
            currNode = treeStack.pop()
    }
}

Another iterative approach of preorder using stack

If we simplify the preorder traversal for each node, we process the node first and then process the left and right children. To track this order using stack, we push right child before left child to ensure that left subtree is processed first. In other words: If we pop a node from stack, the left child comes before the right child.

Implementation steps

  1. We create empty stack treeStack and push root node.
  2. We also use pointer currNode to track the current node.
  3. Now we run a loop till treeStack is not empty.
  4. We pop the top node of stack and assign it with currNode. Now we move forward to process the currNode.
  5. if(currNode->right != NULL), we push the right child of currNode.
  6. if(currNode->left != NULL), we push the left child of currNode.
  7. Now we move to the next iteration of loop.

Pseudocode of the Implementation

void iterativePreorder(TreeNode root) 
{
    if(root == NULL)
        return
    Stack<TreeNode> treeStack
    TreeNode currNode   
    treeStack.push(root)
    while (treeStack.empty() == false) 
    {
        currNode = treeStack.pop()
        process(currNode->data)
        if (currNode->right != NULL)
            treeStack.push(currNode->right)
        if (currNode->left != NULL)
            treeStack.push(currNode->left)
    }
}

Analysis of iterative preorder traversal

Each node is pushed into and popped out of the treeStack only once. So we are doing constant number of operations for each node. Time complexity = n* O(1) = O(n).

Space complexity: We are using one stack and the size of stack depends on the height of binary tree. Think! So space complexity = O(h).

Iterative inorder traversal  using stack

In inorder traversal, we first process the left subtree, then root node, and finally the right subtree. How do we simulate this process using stack? Let’s understand the flow of recursion.

  • We first start from the root and go to the left subtree, then to the root of left subtree, and so on ... we reach the leftmost end of tree. In such a way, recursion first processes the leftmost node.
  • If the leftmost node is a node with right child only: Recursion goes to the right child of the node. After processing all nodes in the right subtree using the same process, recursion backtrack to the node. Now traversal of the leftmost node is done, and recursion backtrack to its parent node. A similar process continues for the parent node.
  • If the leftmost node is a leaf node: Then traversal of the leftmost node is done, and recursion backtrack to its parent node. Now it processes the parent node, goes to the right subtree of the parent node, processes all nodes in the right subtree, and backtracks to the parent node. This process continues further in a similar way.

Inorder traversal example

Here is an idea of iterative simulation using stack: Before processing the left subtree of any node, we need to save that node on stack (to process that node and right subtree of that node later). Then we go to the left child of that node. After processing all nodes in the left subtree, we pop node from the top of stack, process it, and go to the right child of that node to traverse the right subtree. Let's understand it via the clear implementation steps.

Implantation steps  of Iterative Inorder using stack

  • We create an empty stack treeStack and push the root node.
  • We also declare pointer currNode to track the current node.
  • Now we run a loop till treeStack is not empty or currNode != NULL

Condition 1: If (currNode != NULL)

We push currNode into stack for processing currNode and right subtree later. Now we move to the left child i.e. currNode = currNode->left and push it into the stack. This step will continue in a loop till we reach the leftmost end of binary tree i.e. currNode == Null.

if (currNode != NULL) 
{
    treeStack.push(currNode)
    currNode = currNode->left
}

Condition 2: If (currNode == NULL)

Now we reached the leftmost end of binary tree, so we move to the parent of that node by popping a node from the top of treeStack. We assign it with currNode, and process it. At this stage, we need to traverse the right subtree of parent node, so we assign currNode with the right child of popped node and continue the above process in loop. Think!

if (currNode == NULL) 
{
    currNode = treeStack.pop()
    process(currNode->data)
    currNode = currNode->right
}

Pseudo code of iterative inorder traversal

void iterativeInorder(TreeNode root) 
{
    if(root == NULL) 
        return
    Stack<TreeNode> treeStack
    TreeNode currNode = root
    while (treeStack.empty() == false || currNode != NULL) 
    {
        if (currNode != NULL) 
        {
            treeStack.push(currNode)
            currNode = currNode->left
        }
        else 
        {
            currNode = treeStack.pop()
            process(currNode->data)
            currNode = currNode->right
        }
    }
}

Analysis of iterative inorder traversal

Each node is pushed into and popped out of treeStack only once. So we are doing constant number of operations for each node. Time complexity = n* O(1) = O(n).

Space complexity: We are using one stack and the size of stack depends on the height of binary tree. Think! So space complexity = O(h).

Iterative postorder traversal  using stack

In postorder traversal, we first process left subtree, then we process right subtree, and finally process the root node. In other words, we process all nodes in left and right subtree before processing any node. How do we simulate this process using stack? Let’s track the flow of recursion.

  • We first start from the root and go to left subtree, then go to the root of left subtree, and so on...we reach the leftmost end of binary tree.
  • If the leftmost node is a node with right child only: Recursion goes to the right child of that node. After processing all nodes in the right subtree, recursion backtrack to that node and process it. Now traversal of the leftmost node is done, and recursion backtrack to its parent node. From here, recursion goes to the right child of the parent node, processes all nodes in the right subtree, backtracks, and finally processes the parent node. This process continues similarly.
  • If the leftmost node is a leaf node: From here, recursion process leftmost node and traversal of the leftmost node is done. Now recursion backtracks to its parent node. From here, recursion goes to the right child, processes all nodes in the right subtree of parent node, backtracks to the parent node, and finally processes the parent node. This process continues similarly.

Postorder traversal example

Here is an idea of iterative simulation using stack: Before processing the left subtree of any node, we need to save two things on the stack: 1) Right child of the current node to process right subtree after the traversal of left subtree 2) Current node, so that we can process it after the traversal of right subtree. To simulate this, we can use two separate stacks to store these two nodes. But how do we track the nodes in a postorder fashion? Let’s understand it clearly via the implementation.

Implantation steps : Iterative postorder using two stacks

  • We declare two separate stacks: rightChildStack to store the right child and mainStack to store the current node. We also initialize an extra pointer currNode to track the current node during the traversal.
  • Now we run a loop till mainStack is not empty or currNode != NULL. 

Condition 1: If (currNode != NULL)

If the right child of the currNode is not NULL, then we push the right child into the rightChildStack. Now we push the currNode into the mainStack and go to the left child. We continue the same process until we reach the leftmost end of the tree i.e currNode == NULL.

if(currNode != NULL) 
{
    if(currNode->right != NULL)
        rightChildStack.push(currNode->right)
    mainStack.push(currNode)
    currNode = currNode->left
}

Condition 2: If (currNode == NULL)

Now we access the parent node of the currNode which is at the top of the mainStack. We access the top node and assign it with currNode. But before processing the currNode, we need to first process nodes in the right subtree. 

  • If the rightChildStack is not empty and the top node of the rightChildStack is equal to the right child of the currNode, it means, we have not yet traversed the subtree rooted at the right child. So we pop the top node from the rightChildStack and assign it with currNode. 
  • Otherwise, traversal of the right subtree of the currNode is done, and we finally process the currNode. After this, we pop the top node from the mainStack and update currNode with NULL (Think). In this case, the loop again goes to condition 2 and repeats the same steps until currNode is not equal to NULL.
currNode = mainStack.top()
if(!rightChildStack.isEmpty() && currNode->right == rightChildStack.top())
    currNode = rightChildStack.pop()
else 
{
    process(currNode->data)
    mainStack.pop()
    currNode = NULL
}

Pseudocode of the iterative postorder

void iterativePostorder (TreeNode root) 
{
    if(root == null) 
        return
    Stack<TreeNode> mainStack
    Stack<TreeNode> rightChildStack
    TreeNode currNode = root
    while(!mainStack.empty() || currNode != NULL) 
    {
        if(currNode != NULL) 
        {
            if(currNode->right != NULL)
                rightChildStack.push(currNode->right)
            mainStack.push(currNode)
            currNode = currNode->left
        }
        else 
        {
            currNode = mainStack.top()
            if(!rightChildStack.isEmpty() && currNode->right == rightChildStack.top())
                currNode = rightChildStack.pop()
            else 
            {
                process(currNode->data)
                mainStack.pop()
                currNode = NULL
            }
        }
    }
}

Analysis of iterative postorder

Each node is pushed into and popped out of the mainStack only once. Similarly, in the worst case, each node is pushed into the rightChildStack at most once. So we are doing a constant number of operations for each node. Time complexity = n* O(1) = O(n).

Space complexity: We are using two different stacks where the size of each stack depends on the height of the binary tree. Think! So the space complexity = O(h).

Critical ideas to think!

  • Can we optimize the above postorder traversal? Is there any way to implement the post-order traversal using a single stack? Here is an implementation code using the one stack and O(n) extra memory. Is it look similar to the two-stack implementation? Think!

    void iterativePostorderUsingSet(TreeNode root) 
    {
      if(root == null)
          return
      Stack<TreeNode> treeStack
      HashSet<TreeNode> visitedNodeSet
      TreeNode currNode = root
      while(!treeStack.empty() || currNode != NULL) 
      {
          if(currNode != NULL) 
          {
              if(visitedNodeSet.contains(currNode)) 
              {
                  process(currNode)
                  currNode = NULL
              } 
              else 
              {
                  treeStack.push(currNode)
                  if(currNode->right != NULL)
                      treeStack.push(currNode->right)
                  visitedNodeSet.add(currNode)
                  currNode = currNode->left
              }
          }
          else
              currNode = treeStack.pop()
      }
    }
  • How do we implement all the iterative DFS traversals without using stack? Idea: Threaded binary tree traversal or Morris Traversal.
  • What would be the worst, best, and average case of space complexity?
  • In the postorder traversal, how many operations will be performed on rightChildStack in the best and worst case?

Critical concepts to explore further!

Problems to solve using DFS traversals

  • Left View of Binary Tree
  • Maximum width of a binary tree
  • Min Depth of Binary Tree
  • Level with a maximum number of nodes
  • Convert a binary tree into a mirror tree
  • Find the diameter of the binary tree
  • Print nodes at k distance from a given node
  • Boundary Traversal of the binary Tree
  • Least Common Ancestor of two nodes in Binary tree
  • Construct binary tree from preorder and In-order Traversal

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