2. Maximum Depth of Binary Tree

Description

Maximum Depth of Binary Tree
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, 104].
• -100 <= Node.val <= 100

Approach - BFS

• We use queue which uses linked lists, we start by adding the root node then go BFS
• We check how long it takes to reach the end
• Time: O(n) Space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        if (root != null) q.add(root); //tp kickstart
        int l = 0;
        while (!q.isEmpty()) {
            int size = q.size(); //use this instead direclty in for loop because the queue size will be altered inside the loop
            for (int i = 0; i < size; i++) {
                TreeNode n = q.poll(); // npt only removes but returns
                if (n.left != null) q.add(n.left);
                if (n.right != null) q.add(n.right);
            }
            l++;
        }
        return l;
    }
}

Approach - Recursion

• Time complexity: O(n)
• Space complexity: O(h)
  • Best Case (balanced tree): O(log(n))
  • Worst Case (degenerate tree): O(n)

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

Approach 1: Recursive DFS ($O(N)$ Time, $O(N)$ Space)

Intuition

The maximum depth of a binary tree is $1$ plus the maximum depth between its left and right subtrees.

1. If the current node is null, return $0$ (base case).
2. Recursively calculate the depth of root.left and root.right.
3. Return 1 + Math.max(leftDepth, rightDepth).

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

Complexity

• Time Complexity: $O(N)$ — Visits every node in the binary tree exactly once.
• Space Complexity: $O(N)$ worst-case call stack depth for a skewed tree ($O(\log N)$ for a balanced tree).

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Approach 2: BFS Level-Order Traversal ($O(N)$ Time, $O(N)$ Space)

Intuition

Traverse the tree level by level using a queue (ArrayDeque).

1. Add root to the queue before the loop.
2. For each iteration of the while (!queue.isEmpty()) loop, increment a depth counter.
3. Process all nodes at the current level (levelSize = queue.size()) and push their non-null children into the queue.
4. Return depth once the queue is empty.

import java.util.ArrayDeque;
import java.util.Deque;
 
class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) return 0;
 
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        int depth = 0;
 
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            depth++;
 
            for (int i = 0; i < levelSize; i++) {
                TreeNode curr = queue.poll();
 
                if (curr.left != null) queue.offer(curr.left);
                if (curr.right != null) queue.offer(curr.right);
            }
        }
 
        return depth;
    }
}
 

Complexity

• Time Complexity: $O(N)$ — Every node is offered and polled from the queue once.
• Space Complexity: $O(N)$ — Maximum queue size is bounded by the widest level of the tree (up to $N/2$ nodes).

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Easy Memory Rule

“DFS: 1 + Math.max(maxDepth(left), maxDepth(right)) $\to$ OR BFS: Increment depth once per level in standard queue loop!”