Maximum Depth of Binary Tree

LeetCode 104 | Difficulty: Easy

Easy

Problem Description

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Example 1:

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

Example 2:

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

Constraints:

- The number of nodes in the tree is in the range `[0, 10^4]`.

- `-100 <= Node.val <= 100`

Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Tree BFS (Level-Order)

Use a queue to process the tree level by level. At each level, process all nodes in the queue, then add their children. Track the level size to know when one level ends and the next begins.

When to use

Level-order traversal, level-based aggregation, right/left side view.

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Solutions

Solution 1: C# (Best: 100 ms)

Metric	Value
Runtime	100 ms
Memory	24.5 MB
Date	2019-12-15

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int MaxDepth(TreeNode root) {
        if(root==null)
        {
            return 0;
        }
        else return Math.Max(MaxDepth(root.left), MaxDepth(root.right))+1;
    }
}

📜 3 more C# submission(s)

Submission (2018-05-03) — 108 ms, N/A

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int MaxDepth(TreeNode root) {
        if(root==null)
        {
            return 0;
        }
        else return Math.Max(MaxDepth(root.left), MaxDepth(root.right))+1;
    }
}

Submission (2022-02-14) — 137 ms, 37.8 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int MaxDepth(TreeNode root) {
        if(root==null)
        {
            return 0;
        }
        else return Math.Max(MaxDepth(root.left), MaxDepth(root.right))+1;
    }
}

Submission (2017-12-06) — 149 ms, N/A

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int MaxDepth(TreeNode root) {
        if(root == null) return 0;
        else return (1 + Math.Max(MaxDepth(root.left),MaxDepth(root.right)));
    }
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.