13. Balanced Binary Tree
easyAsked at InstacartCheck if a binary tree is height-balanced — Instacart screens recursion-with-bailout fluency before harder routing-tree balance checks.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given a binary tree, determine if it is height-balanced — that is, for every node, the depth difference between its left and right subtrees is at most one.
Constraints
The number of nodes in the tree is in the range [0, 5000]-10^4 <= Node.val <= 10^4
Examples
Example 1
root = [3,9,20,null,null,15,7]trueExample 2
root = [1,2,2,3,3,null,null,4,4]falseApproaches
1. Top-down depth check
Compute depth at every node and compare children.
- Time
- O(n^2)
- Space
- O(h)
function depth(n) { return n ? 1 + Math.max(depth(n.left), depth(n.right)) : 0; }
function check(n) { if (!n) return true; return Math.abs(depth(n.left)-depth(n.right)) <= 1 && check(n.left) && check(n.right); }Tradeoff:
2. Bottom-up with sentinel
Return -1 from a recursive helper as soon as imbalance is found.
- Time
- O(n)
- Space
- O(h)
function isBalanced(root) {
function 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:
Instacart-specific tips
Instacart wants the single-pass version — they'll ask how this would scale if applied to a delivery-zone hierarchy of 10M nodes.
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