Binary Tree Preorder Traversal - Problem
Given the root of a binary tree, return the preorder traversal of its nodes' values.
Preorder traversal follows the pattern: Root → Left → Right. This means we visit the current node first, then recursively traverse the left subtree, followed by the right subtree.
For example, in a tree with root value 1, left child 2, and right child 3, the preorder traversal would be [1, 2, 3].
Note: You may implement this using recursion or iteration. Both approaches are valuable to understand!
Input & Output
example_1.py — Basic Tree
$
Input:
root = [1, null, 2, 3]
›
Output:
[1, 2, 3]
💡 Note:
Start at root (1), no left child so go to right child (2). At node 2, visit left child (3) first, then right (null). Final order: 1 → 2 → 3
example_2.py — Empty Tree
$
Input:
root = []
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Output:
[]
💡 Note:
An empty tree has no nodes to traverse, so return an empty list
example_3.py — Single Node
$
Input:
root = [1]
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Output:
[1]
💡 Note:
Tree with only root node returns just that single value
Constraints
-
The number of nodes in the tree is in the range
[0, 100] -
-100 ≤ Node.val ≤ 100 - Follow up: Recursive solution is trivial, could you do it iteratively?
Visualization
Tap to expand
Understanding the Visualization
1
Visit Current Node
Process/record the current node's value immediately upon arrival
2
Explore Left Branch
Recursively traverse the entire left subtree before moving right
3
Explore Right Branch
After completing left subtree, recursively traverse the right subtree
4
Backtrack
Return to parent node and continue with its remaining subtrees
Key Takeaway
🎯 Key Insight: Preorder traversal naturally follows a "process first, then recurse" pattern, making it perfect for operations like copying trees or prefix expression evaluation.
💡
Explanation
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