DSA interview question

Binary Tree Postorder Traversal

Easy · Data structures · Topics: tree, dfs, recursion

The problem

A disk-usage tool computes the size of every folder, and a folder's size is only known after all its subfolders have been measured. Processing children before their parent is postorder: left subtree, right subtree, then the node. The same pattern appears when deleting a tree (free the children before the parent) and when evaluating expression trees (evaluate the operands before applying the operator).

Example: [1, null, 2, 3] → [3, 2, 1].

Function signature

vector<int> postorderTraversal(TreeNode* root);

Test cases

Input (level order)	Output	Explanation
`[1, null, 2, 3]`	`[3, 2, 1]`
`[1, 2, 3, 4, 5, null, 8, null, null, 6, 7, 9]`	`[4, 6, 7, 5, 2, 9, 8, 3, 1]`
`[2, 1, 3]`	`[1, 3, 2]`
`[]`	`[]`
`[1]`	`[1]`

(Confirm: LeetCode's level-order input format; the tree may be empty.)

How interviewers phrase it

"Given the root of a binary tree, return the postorder traversal of its nodes' values" (LeetCode 145).

Clarifying questions to ask

Recursive or iterative?
How deep can the tree be?
What should an empty tree return? (An empty list.)

Follow-ups to expect

"Why is postorder the natural order for deletion or size computation?"
"Iterative with two stacks?"
"Iterative with one stack?"
"How does postorder relate to preorder?"
"Complexity?"

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