DSA interview question

Diameter of a Binary Tree

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

The problem

A network of routers is wired as a binary tree, and the engineers want the worst-case hop count between any two routers: the diameter, the number of edges on the longest path between any two nodes. That path doesn't have to pass through the root. For every node, the longest path that "bends" there is height(left) + height(right) (in edges). So a single post-order pass that computes heights can also track the largest such sum.

Example: [1, 2, 3, 4, 5] → 3 (path 4 → 2 → 1 → 3).

Function signature

int diameterOfBinaryTree(TreeNode* root);  // number of edges

Test cases

Input (level order)	Output	Explanation
`[1, 2, 3, 4, 5]`	`3`	4-2-1-3 or 5-2-1-3
`[1, 2, null, 3, 4, 5, null, null, 6]`	`4`	5-3-2-4-6, which avoids the root
`[1, 2]`	`1`
`[1]`	`0`	A single node: no edges
`[]`	`0`

(Confirm: the length is counted in edges; the path can start and end anywhere.)

How interviewers phrase it

"Given the root of a binary tree, return the length of the diameter of the tree: the longest path between any two nodes" (LeetCode 543).

Clarifying questions to ask

Edges or nodes? (Edges.)
Must the path go through the root? (No.)
Can the tree be empty? (Return 0.)

Follow-ups to expect

"Why can't you just use the root?"
"What does each node contribute?"
"Why is it O(n)?"
"How would you do it for a general tree or graph?"
"Complexity?"

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