Skip to main content

13. Balanced Binary Tree

easyAsked at Unity

Decide if a tree is height-balanced. Unity uses this for BVH-balance checks during physics broadphase rebuilds.

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

Problem

Given a binary tree, determine if it is height-balanced: at every node the heights of left and right subtrees differ by at most 1.

Constraints

  • 0 <= nodes <= 5000
  • -10^4 <= node.val <= 10^4

Examples

Example 1

Input
root=[3,9,20,null,null,15,7]
Output
true

Example 2

Input
root=[1,2,2,3,3,null,null,4,4]
Output
false

Approaches

1. Compute depth at every node

For each node compute left/right depths separately, leading to repeated work.

Time
O(n^2)
Space
O(h)
const depth = n => n ? 1+Math.max(depth(n.left),depth(n.right)) : 0;
function isBal(n) { return !n || (Math.abs(depth(n.left)-depth(n.right))<=1 && isBal(n.left) && isBal(n.right)); }

Tradeoff:

2. DFS sentinel

Return -1 once any subtree is unbalanced; otherwise return the height. Single traversal.

Time
O(n)
Space
O(h)
function isBalanced(root) {
  const dfs = (n) => {
    if (!n) return 0;
    const l = dfs(n.left); if (l === -1) return -1;
    const r = dfs(n.right); if (r === -1) return -1;
    if (Math.abs(l - r) > 1) return -1;
    return 1 + Math.max(l, r);
  };
  return dfs(root) !== -1;
}

Tradeoff:

Unity-specific tips

Unity grades for single-pass DFS because BVH balance checks during physics broadphase can't afford repeated subtree walks.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

Practice these live with InterviewChamp.AI

Drill Balanced Binary Tree and other Unity interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →