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10. Same Tree

easyAsked at Figma

Given the roots of two binary trees, determine whether they are structurally identical and have the same values. Figma uses this to check tree-diffing fluency, the foundation of their multiplayer scene-graph reconciliation step.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Figma loops.

  • Glassdoor (2026-Q1)— Figma onsite, paired with multiplayer-reconciliation discussion.
  • LeetCode Discuss (2025)— Easy bucket of Figma OA.

Problem

Given the roots of two binary trees p and q, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

Constraints

  • The number of nodes in both trees is in the range [0, 100].
  • -10^4 <= Node.val <= 10^4

Examples

Example 1

Input
p = [1,2,3], q = [1,2,3]
Output
true

Example 2

Input
p = [1,2], q = [1,null,2]
Output
false

Approaches

1. Serialize and compare strings

Serialize both trees with a deterministic null-marker scheme; compare strings.

Time
O(n)
Space
O(n)
function isSameTree(p, q) {
  const ser = (n) => n ? `(${n.val},${ser(n.left)},${ser(n.right)})` : '#';
  return ser(p) === ser(q);
}

Tradeoff: Linear but allocates a giant string. Hides the structural recursion that interviewers want to see.

2. Recursive structural compare

Both null => true; one null => false; values differ => false; else recurse on left and right.

Time
O(n)
Space
O(h)
function isSameTree(p, q) {
  if (!p && !q) return true;
  if (!p || !q) return false;
  if (p.val !== q.val) return false;
  return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}

Tradeoff: Clean, expressive, and short-circuits on the first mismatch. Best for showing recursive thinking.

Figma-specific tips

Figma loves the recursive form because their multiplayer reconciler walks two scene graphs in lockstep doing exactly this comparison. Lead with the three-case taxonomy ('both null / one null / both present') aloud, and explicitly mention that the && short-circuits on the first mismatch — that's the Figma-style observation about avoiding unnecessary tree walks under load.

Common mistakes

  • Checking p.val !== q.val before the null guards — null pointer exception.
  • Returning early on the left subtree without checking the right when both match the value.
  • Using == instead of === for value comparison (rarely matters here but signals carelessness).

Follow-up questions

An interviewer at Figma may pivot to one of these next:

  • Symmetric Tree (LC 101) — same skeleton but compares left/right mirror.
  • Subtree of Another Tree (LC 572) — uses isSameTree as a primitive.
  • Tree edit distance — generalize to a diff cost.

Solve it now

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Output

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FAQ

What if both trees are empty?

They're considered the same. The first base case (!p && !q => true) handles it.

Could I use BFS instead?

Yes — walk both with the same queue order. DFS recursion is shorter and stack-safe up to ~10^4 depth which fits this problem.

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