2. 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));
    }
}