[GeeksForGeeks] Find the maximum depth or height of a tree

Given a binary tree, find the maximum depth or height of this tree.

Solution 1. In order traversal to count the maximum depth

import java.util.LinkedList;
import java.util.Queue;

class TreeNode {
    TreeNode left;
    TreeNode right;
    int val;
    TreeNode(int val){
        this.left = null;
        this.right = null;
        this.val = val;
    }
}
public class Solution {
    public int findMaxDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        int maxDepth = 0;
        queue.offer(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode curr = queue.poll();
                if(curr.left != null){
                    queue.offer(curr.left);
                }
                if(curr.right != null){
                    queue.offer(curr.right);
                }
            }
            maxDepth++;
        }
        return maxDepth;
    }
}

Solution 2. Recursion on both left and right subtree.

class TreeNode {
    TreeNode left;
    TreeNode right;
    int val;
    TreeNode(int val){
        this.left = null;
        this.right = null;
        this.val = val;
    }
}
public class Solution {
    public int findMaxDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        int leftDepth = findMaxDepth(root.left);
        int rightDepth = findMaxDepth(root.right);
        return leftDepth > rightDepth ? 1 + leftDepth : 1 + rightDepth;
    }
}

Both solution 1 and 2's runtime are O(n), n is the total number of nodes in the given binary tree. 

Solution 1 uses O(n) extra memory for the queue used, where solution 2 only uses O(1) memory 

for each recursive call. But the downside of the recursive solution is the the methods stack may

get very deep.