Balanced Binary Tree

easycodingAlgorithmsBinary TreesDFS~15 min

Asked at Google, Meta

Overview

Given a binary tree, determine if it is height-balanced — a tree where the depth of the two subtrees of every node never differs by more than 1.

Constraints

The number of nodes is in the range [0, 5000].
-10^4 <= Node.val <= 10^4
The tree is given as an array in level-order (null for missing nodes).

Examples

isBalanced([3, 9, 20, null, null, 15, 7]);
// => true

isBalanced([1, 2, 2, 3, 3, null, null, 4, 4]);
// => false

Solution

Reveal solution

function isBalanced(root) {
  function height(node) {
    if (!node) return 0;
    const left = height(node.left);
    if (left === -1) return -1;
    const right = height(node.right);
    if (right === -1) return -1;
    if (Math.abs(left - right) > 1) return -1;
    return Math.max(left, right) + 1;
  }
  return height(root) !== -1;
}

Resources

Balanced Binary Tree — LeetCode

balanced-binary-tree.js

Balanced Binary Tree

easycodingAlgorithmsBinary Trees

Overview

Given a binary tree, determine if it is height-balanced — a tree where the depth of the two subtrees of every node never differs by more than 1.

Constraints

The number of nodes is in the range [0, 5000].
-10^4 <= Node.val <= 10^4
The tree is given as an array in level-order (null for missing nodes).

Examples

isBalanced([3, 9, 20, null, null, 15, 7]);
// => true

isBalanced([1, 2, 2, 3, 3, null, null, 4, 4]);
// => false

Solution

Reveal solution

function isBalanced(root) {
  function height(node) {
    if (!node) return 0;
    const left = height(node.left);
    if (left === -1) return -1;
    const right = height(node.right);
    if (right === -1) return -1;
    if (Math.abs(left - right) > 1) return -1;
    return Math.max(left, right) + 1;
  }
  return height(root) !== -1;
}

Resources

Balanced Binary Tree — LeetCode

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