What are Inorder, Preorder, and Postorder Tree Traversals?
Learn the difference between inorder, preorder, and postorder tree traversals, when to use each, and how to answer this interview question.
Expected Interview Answer
Inorder, preorder, and postorder are the three depth-first orders for visiting a binary tree's nodes, differing only in when the current node is visited relative to its left and right subtrees, each running in O(n) time and O(h) space for a tree of height h.
Inorder visits left subtree, then the node, then right subtree, which produces sorted order for a binary search tree, making it the go-to traversal for extracting values in order. Preorder visits the node first, then left, then right, which is useful for copying or serializing a tree since the root is processed before its children. Postorder visits left, then right, then the node last, which is essential when children must be fully processed before the parent, such as deleting a tree bottom-up or evaluating an expression tree. All three are typically implemented recursively using the call stack, or iteratively with an explicit stack, and all visit every node exactly once, giving identical O(n) time complexity that differs only in the resulting visit order and its downstream use.
- Inorder yields sorted output on a binary search tree
- Preorder mirrors tree structure, ideal for copying and serialization
- Postorder processes children before parents, ideal for safe deletion
- All run in O(n) time, O(h) space, recursive or iterative
AI Mentor Explanation
Inorder is like reading a knockout bracket's left half of results, then announcing the current match's winner, then reading the right half — which naturally lists teams in seed order if the bracket is built that way. Preorder is like a commentator announcing the current match first, then diving into the left bracket's results, then the right bracket's, mirroring how the tournament tree itself was structured. Postorder is like only being able to announce a round's winner after both the left and right sub-brackets have fully finished, exactly how a champion can only be crowned once every earlier match has been decided. All three visit every match exactly once, just in a different announcement order.
Step-by-Step Explanation
Step 1
Inorder: left, node, right
Recurse left, visit the current node, then recurse right — yields sorted order on a BST.
Step 2
Preorder: node, left, right
Visit the current node first, then recurse left, then right — mirrors the tree's own structure, good for serialization.
Step 3
Postorder: left, right, node
Recurse left, then right, then visit the current node last — good for deletion and expression evaluation.
Step 4
Implement recursively or iteratively
Use the call stack for simple recursion, or an explicit stack for an iterative version with O(h) space either way.
What Interviewer Expects
- State the visit order precisely for all three: left/node/right, node/left/right, left/right/node
- Connect inorder on a BST to producing sorted output
- Give a correct use case for preorder (serialization/copying) and postorder (deletion/expression eval)
- State O(n) time and O(h) space for all three, recursive or iterative
Common Mistakes
- Mixing up preorder and postorder visit positions
- Forgetting inorder only yields sorted order specifically for a binary search tree
- Claiming O(1) space for the recursive version, ignoring the call stack
- Not knowing an iterative version needs an explicit stack to match recursive behavior
Best Answer (HR Friendly)
“Inorder, preorder, and postorder are just three different orders for visiting a tree's nodes, based on whether I look at the current node before, in between, or after visiting its children. I use inorder to get sorted values out of a binary search tree, preorder when I need to copy or serialize a tree, and postorder when children must be handled before their parent, like deleting a tree safely.”
Code Example
class TreeNode:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def inorder(node, out=None):
if out is None:
out = []
if node:
inorder(node.left, out)
out.append(node.val)
inorder(node.right, out)
return out
def preorder(node, out=None):
if out is None:
out = []
if node:
out.append(node.val)
preorder(node.left, out)
preorder(node.right, out)
return out
def postorder(node, out=None):
if out is None:
out = []
if node:
postorder(node.left, out)
postorder(node.right, out)
out.append(node.val)
return outFollow-up Questions
- How would you implement inorder traversal iteratively using an explicit stack?
- Why does inorder traversal produce sorted output specifically for a binary search tree?
- How would you reconstruct a binary tree from its preorder and inorder traversals?
- Why is postorder the correct choice for safely deleting a tree?
MCQ Practice
1. Which traversal order visits the current node before its left and right subtrees?
Preorder visits node, then left, then right — the node comes first.
2. Inorder traversal of a binary search tree produces what kind of output?
Because left values are smaller and right values are larger in a BST, left-node-right visiting yields ascending order.
3. Which traversal is most appropriate for safely deleting all nodes in a tree?
Postorder processes both children before the parent, ensuring children are freed before their parent is deleted.
Flash Cards
What is the visit order for inorder traversal? — Left subtree, current node, right subtree.
What is the visit order for preorder traversal? — Current node, left subtree, right subtree.
What is the visit order for postorder traversal? — Left subtree, right subtree, current node.
Why is inorder useful on a binary search tree? — It produces the node values in sorted ascending order.