Intuitive Iterative Binary Tree Traversals

Binary tree traversals are a staple of the technical interview process at many software companies, small and large. For anyone with an understanding of recursion, the family of traversal techniques are quite straightforward. A common twist on these concepts that show up more in technical interviews than undergraduate computer science problem sets is the rather artificial constraint that asks one to implement the traversals using iteration rather than recursion.

I’ve always found the reference implementations of iterative tree traversals, particularly inorder traversal, to be lacking in intuitive understanding. The classic way of iteratively traversing a binary tree is to use a stack data structure, and the first snippet of code you often see is something like this:

stack = []
curr = root
while curr is not null || stack is not empty
    while curr is not null
        curr = curr.left

When I see this code, I immediately have more questions than insight.

  • Why is the code following left pointers while pushing all the nodes onto the stack?
  • Why is the loop condition checking against both the current node and the stack size?
  • Why is there a nested while loop?

To me, the intuitive way to reason about an iterative implementation of a recursive function is to simulate a call stack, and that usually begins with a pen and paper. For example, suppose we have a binary tree that looks like this:

   / \
  B   C
 / \
D   E

An inorder traversal of such a tree should yield the nodes in this order: D, B, E, A, C. The best way to simulate the call stack that yields such a traversal is to draw out the contents of the stack as the traversal makes its way through the tree. I like to model my stacks after the real world, with a physical base to indicate the bottom of the stack, and elements being pushed on and popped off. Here’s the visualization of an empty stack, and its transformation following two push (push(A), push(B)) operations and one pop (pop()) operation:

                  B      ~B~
          A       A       A
_____   _____   _____   _____
stack   stack   stack   stack

At the end of the operations, the stack contains a single element B. The notation here marks any items popped off the stack with strikethrough-like markers (~), but leaves it on the stack in its original location to better illustrate ordering. We can now use this notation to visually simulate the first few steps of what an inorder traversal on the example tree might look like using a standard depth-first search approach. Initially, the stack is empty, and the root node is pushed onto the stack.


We invoke the same logic repeatedly while the stack has items: pop a node off and process it. When a node is popped off the stack, we need to process it in inorder fashion: traverse its left child first, visit itself, then traverse its right child. This translates to push(C), push(A), and push(B). Notice that the elements are pushed onto the stack in reverse order from the way they would be processed in a standard recursive implementation, as to achieve the desired order. Perhaps more importantly, the node itself (A) is pushed back onto the stack.

  A      ~A~     ~A~
_____   _____   _____
stack   stack   stack

At this point, B is popped off the stack and the same logic is applied. The traversal proceeds down to the left child of B, followed by a visit to B, and a subsequent traversal down the right child of B. That is, push(E), push(B), push(D). Here’s what the stack looks like after reaching the first leaf node D:

  B      ~B~     ~B~
  A       A       A
  C       C       C
 ~A~     ~A~     ~A~
_____   _____   _____
stack   stack   stack

Since D has no child nodes, it will be popped off the stack then pushed back onto the stack. Here’s the second piece of logic that is core to our traversal: if a node being processed has already been discovered, then it should be visited. With that, the algorithm for our inorder traversal - with respect to processing a single node - can be expressed as follows:

if the node has already been discovered
    "visit" or do something with the node
    mark the node as discovered
    push the right child of the node onto the stack
    push the node onto the stack
    push the left child of the node onto the stack

As noted earlier, the symmetry between this iterative approach and the standard recursive implementation is clear. The recursive implementation will first (in an eager, depth-first manner) traverse down the left child of a given node, then visit the node, followed by a traversal down the right child. Using an explicit stack data structure to simulate the calling pattern simply means pushing the nodes onto the explicit stack in reverse order as compared to the implicit recursion call stack.

To illustrate the process more clearly, we can push nodes onto the stack with an explicit status: a start status to indicate that the node has yet to be processed and an end status to indicate that it has been processed. For example, A.start and A.end will represent the start and end state for a node A, respectively. Here’s the same visualization for the same first few operations as above, with explicit status attributes for each node:

A.start   ~A.start~   ~A.start~
 _____      _____       _____
 stack      stack       stack

With this extended notation, the logic needed to process any given node at the top of the stack is clear. If a node at the top of the stack has a start status, push its right child onto the stack with a start status, push itself back onto the stack with an end status, and push its left child onto the stack with a start status.

An astute reader will notice that since there are only two states, a binary flag is sufficient for storing the same information. In fact, this can represented in exactly the same way as the classic implementations of graph traversals introduced in CLRS, in which nodes are assigned colors to keep track of traversal progress. In the case of binary tree traversals, we only need two colors: white for undiscovered nodes and black for discovered nodes. With that insight, the iterative version of any of the traversals becomes easy to derive:

def iterative_inorder_traversal(root):
    stack = []
    discovered = {}
    while len(stack) > 0:
        node = stack.pop()
        if node in discovered:
            pass # "Visit" or do something with the node
            discovered[node] = True
            if node.right:
            if node.left:

An iterative implementation of preorder or postorder traversal should easily follow from the inorder traversal; the sequence in which the nodes should be pushed onto the stack simply needs to be modified to match the desired traversal behavior. The cost of the intuitive version of these iterative traversals is a larger constant in the runtime complexity, as each node is actually processed twice. Ultimately, the runtime still grows at a rate linearly proportional to the size of the input.