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3. Same Tree

easyAsked at JetBrains

Check whether two binary trees are structurally and value-equal node-by-node — JetBrains uses this to gauge whether you can compare PSI subtrees for incremental reparse.

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

Problem

Given the roots of two binary trees, return true if they are identical — same shape and same values at every node.

Constraints

  • 0 <= nodes <= 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

Serialize both trees to strings and compare; wasteful allocation for large trees.

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

Tradeoff:

2. Parallel DFS

Recurse both trees in lockstep, short-circuiting on first mismatch. Same shape as JetBrains' AST diff walkers used during incremental parsing.

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:

JetBrains-specific tips

JetBrains expects you to mention this is exactly the shape of incremental-reparse diffing — short-circuit early and avoid building intermediate strings.

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Output

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