Binary Tree Preorder Traversal
An easy Binary Trees problem included in Apna College, Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Binary Trees
- Sheets
- 3
- Core for
- 4 roles
- Platform
- LeetCode
The problem
Given the root of a binary tree, return the preorder traversal of its nodes' values. Preorder traversal visits the root node first, then the left subtree, then the right subtree.
Example 1
- Input
- [1,null,2,3]
- Output
- [1,2,3]
- Why
- Visit root 1 first, then go left (null), then go right to 2 and its left child 3. Order: 1,2,3.
Example 2
- Input
- []
- Output
- []
- Why
- An empty tree produces an empty traversal.
Example 3
- Input
- [1,2]
- Output
- [1,2]
- Why
- Root 1 is visited first, then its left child 2.
Constraints
- The number of nodes is in the range [0, 100].
- -100 <= Node.val <= 100
How to think about it
Updated 2026-09-09Preorder processes each node the instant your eyes first land on it. Unlike inorder or postorder, no state needs to be suspended waiting for descendants; consumption is immediate, meaning any iterative stack has to remember where to resume next, pushing right before left so left pops first.
Approaches, worst first
Recursive visit
time O(n) · space O(n)
Emit the current node value, recurse on the left child, then recurse on the right. Clean and direct, but retains an implicit stack proportional to tree depth.
Iterative stack with child ordering
time O(n) · space O(n)
Pop the current node, push its right child, then push its left child. Because the stack is last-in-first-out, the left subtree is processed entirely before the right subtree begins.
Morris threadingWrite this one
time O(n) · space O(1)
Connect the inorder predecessor's right pointer to the current node, but emit the value the first time the node is linked rather than when the link is broken. Traversal runs in place without extra space.
Where people lose marks · 3
- Pushing the left child before the right child onto an explicit stack reverses the traversal order to root-right-left.
- Pushing null children into the stack without filtering causes unnecessary pop cycles and potential null dereferences.
- Omitting the empty-tree check at the very top causes an immediate undefined error when pushing root to an iterative stack.
The theory behind it
Binary Trees — the ground this problem stands on. All Binary Trees problems
What Binary Trees is
A binary tree is a branching data structure that starts at a single top node called the root, like an upside-down family tree. Every node holds a piece of data and can branch out to at most two children below it, known as the left child and the right child. Because there is no ordering rule about which values go left or right, finding a specific item can require checking every single node in the entire tree.
When to reach for it
Reach for binary trees when problems present hierarchical data with left and right child pointers. Questions asking for tree height, maximum depth, path sums from root to leaf, diameter, lowest common ancestor, or checking whether two trees are mirror reflections of each other all signal binary tree traversals. Any problem asking to inspect or reconstruct a tree layer by layer or path by path belongs here.
How the pattern works
Think recursively by focusing on what a single node must do. If the current node is null, return the base answer immediately. Otherwise, ask the left child for its result, ask the right child for its result, and combine both answers with the current node value before returning up to the parent. For horizontal scans, use a queue to read nodes layer by layer, measuring the queue length at the start of each layer to group nodes by depth.
What each operation costs
| Operation | Time |
|---|---|
| traverse all nodes using recursion or queue | O(n) |
| search for an arbitrary value in an unordered tree | O(n) |
| call stack memory on balanced tree | O(log n) |
| call stack memory on skewed tree | O(n) |
What usually goes wrong with Binary Trees
- Dereferencing left or right child pointers without checking if the current node is null, throwing null pointer errors on empty trees or leaf nodes.
- Defining a leaf node incorrectly by stopping when either child is null instead of checking that both left and right children are simultaneously null.
- Computing tree diameter by taking left height plus right height inside a recursive helper without updating a global maximum across every visited node.
Which roles need this problem
Binary Trees is a core topic for these 4 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
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