Diameter of Binary Tree – Solution & Complexity

Solution Walkthrough

1. Recognize the Pattern

The key pattern is: Bottom-up DFS height computation that updates a running best at every node. Identify the state and invariant before coding.

2. Build the Algorithm

Advance one state transition at a time. Mark or update state before exploring dependent work.

3. Check Edge Cases

Test empty or minimal input, skewed shapes, duplicates where allowed, and impossible outcomes.

4. Solution and Complexity

Time: O(n). Space: O(h) for the recursion stack.

All 5 languages below run and submit against the remote judge for this problem — Java/Go/Rust use the same real TreeNode structural-type support the judge added for the rest of the linked-list/tree track.

def diameter_of_binary_tree(root: TreeNode) -> int:
    best = 0

    def depth(node):
        nonlocal best
        if node is None:
            return 0
        left = depth(node.left)
        right = depth(node.right)
        best = max(best, left + right)
        return 1 + max(left, right)

    depth(root)
    return best

FAQ