Coding problem · Easy · Trees

Maximum Depth of Binary Tree

Easy Algorithms & data structures Seen at Amazon, LinkedIn, Uber

Given the root of a binary tree, return its maximum depth — the number of nodes along the longest path from the root down to the farthest leaf. Trees are shown in level order with null for missing children.

Example 1

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

Example 2

Input: root = [1,null,2]
Output: 2

Hints

Hint 1

depth(node) = 1 + max(depth(left), depth(right)); an empty tree has depth 0.

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Solution and explanation

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Time O(n) · Space O(h) recursion depth

Recursive DFS returns 1 + max(left, right). For very deep (skewed) trees, an iterative BFS counting levels avoids stack overflow.

Reference solution

function maxDepth(root) {
  if (!root) return 0;
  return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}

function maxDepth(root: TreeNode | null): number {
  return root ? 1 + Math.max(maxDepth(root.left), maxDepth(root.right)) : 0;
}

def max_depth(root: Optional[TreeNode]) -> int:
    if root is None:
        return 0
    return 1 + max(max_depth(root.left), max_depth(root.right))

<?php

function maxDepth(?TreeNode $root): int
{
    return $root === null ? 0 : 1 + max(maxDepth($root->left), maxDepth($root->right));
}

class Solution {
    public int maxDepth(TreeNode root) {
        return root == null ? 0 : 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
    }
}

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