Tree DFS - Post-order Traversal
Description
Given the root of a binary tree, return a list of node values in
post-order: left subtree, then right subtree, then the node itself.
Example:
1
/ \
2 3
/ \
4 5
Post-order: [4, 5, 2, 3, 1] Constraints
- 0 <= number of nodes <= 1000 - Node values are integers - Return [] for an empty tree
solution.py
output
Run the cases to see results here.