Check if Binary Tree is Balanced in DSA Typescript

Mental Model

A balanced binary tree means no branch is much longer than another. We check if every node's left and right branches differ in height by at most one.

Analogy: Imagine a tree with branches. If one branch is much longer than its sibling, the tree looks lopsided and might fall. We want to make sure all branches grow evenly.

      1
     / \
    2   3
   / 
  4  

Balanced tree example: left and right heights differ by at most 1.

Dry Run Walkthrough

Input: Binary tree: 1 with left child 2, right child 3; 2 has left child 4

Goal: Determine if the tree is balanced by checking height differences at each node

Step 1: Check height of node 4 (leaf)

Node 4 height = 0 (no children)

Why: Leaf nodes have height 0, base case for height calculation

Step 2: Check height of node 2: left height = 0 (node 4), right height = -1 (no right child)

Height difference at node 2 = 1 (0 - (-1))

Why: Difference 1 is allowed, so node 2 is balanced

Step 3: Check height of node 3 (leaf)

Node 3 height = 0

Why: Leaf node height is 0

Step 4: Check height of root node 1: left height = 1 (node 2), right height = 0 (node 3)

Height difference at node 1 = 1 (1 - 0)

Why: Difference 1 is allowed, so root is balanced

Step 5: Since all nodes are balanced, return true

Tree is balanced

Why: No node violates the height difference rule

Result:

1
/ \
2   3
/ 
4

Balanced: true

Annotated Code

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

function isBalanced(root: TreeNode | null): boolean {
  function checkHeight(node: TreeNode | null): number {
    if (node === null) return -1; // base case: empty subtree height

    const leftHeight = checkHeight(node.left); // get left subtree height
    if (leftHeight === Number.MIN_SAFE_INTEGER) return Number.MIN_SAFE_INTEGER; // left subtree unbalanced

    const rightHeight = checkHeight(node.right); // get right subtree height
    if (rightHeight === Number.MIN_SAFE_INTEGER) return Number.MIN_SAFE_INTEGER; // right subtree unbalanced

    if (Math.abs(leftHeight - rightHeight) > 1) return Number.MIN_SAFE_INTEGER; // current node unbalanced

    return Math.max(leftHeight, rightHeight) + 1; // height of current node
  }

  return checkHeight(root) !== Number.MIN_SAFE_INTEGER;
}

// Driver code
const root = new TreeNode(1,
  new TreeNode(2, new TreeNode(4), null),
  new TreeNode(3)
);

console.log(isBalanced(root));

if (node === null) return -1; // base case: empty subtree height

return -1 height for empty subtree to count edges correctly

const leftHeight = checkHeight(node.left);

recursively get left subtree height

if (leftHeight === Number.MIN_SAFE_INTEGER) return Number.MIN_SAFE_INTEGER;

propagate unbalanced signal up if left subtree unbalanced

const rightHeight = checkHeight(node.right);

recursively get right subtree height

if (rightHeight === Number.MIN_SAFE_INTEGER) return Number.MIN_SAFE_INTEGER;

propagate unbalanced signal up if right subtree unbalanced

if (Math.abs(leftHeight - rightHeight) > 1) return Number.MIN_SAFE_INTEGER;

detect unbalanced node by height difference > 1

return Math.max(leftHeight, rightHeight) + 1;

return height of current node as max child height + 1

return checkHeight(root) !== Number.MIN_SAFE_INTEGER;

final check: if unbalanced signal not found, tree is balanced

Output

true

Complexity Analysis

Time: O(n) because we visit each node once to calculate heights and check balance

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Naive approach recalculates heights repeatedly causing O(n^2) time; this method calculates height and balance in one pass for O(n) time

Edge Cases

Empty tree (root is null)

Returns true because empty tree is balanced

DSA Typescript

if (node === null) return -1;

Single node tree

Returns true because single node has no imbalance

DSA Typescript

if (node === null) return -1;

Unbalanced tree with one branch much deeper

Returns false when height difference > 1 detected

DSA Typescript

if (Math.abs(leftHeight - rightHeight) > 1) return Number.MIN_SAFE_INTEGER;

When to Use This Pattern

When you need to check if a tree is balanced, use a bottom-up height check with early stopping to detect imbalance efficiently.

Common Mistakes

Mistake: Calculating height separately from balance check causing repeated work
Fix: Combine height calculation and balance check in one recursive function with early return

Mistake: Returning 0 for null nodes instead of -1, causing off-by-one height errors
Fix: Return -1 for null nodes to count edges correctly in height

Summary

Checks if a binary tree is balanced by ensuring height difference between left and right subtrees is at most one at every node.

Use when you want to verify tree balance to optimize operations like search or insert.

The key insight is to calculate height and check balance in one pass, returning a special value to stop early if unbalanced.