Leetcode (Java) -104. The maximum depth of the binary tree

Given a binary tree to find its maximum depth.

The depth of the binary tree is the root node to the nodes on the longest path farthest leaf node.

Description: leaf node is a node has no child nodes.

Example:
given binary tree [3,9,20, null, null, 15,7 ],

    3
   / \
  920
    / \
   157
returns to its maximum depth 3.

Ideas 1: Recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;//退出条件
        
        int leftMaxDepth = maxDepth(root.left);
        int rightMaxDepth = maxDepth(root.right);

        return Math.max(leftMaxDepth,rightMaxDepth)+1;
    }
}

Ideas 2: BFS-level binary tree traversal

class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;

        LinkedList<TreeNode> queue=new LinkedList<>();
        queue.add(root);
        int maxDepth=0;
        while(!queue.isEmpty()){
            maxDepth++;
            int levelSize=queue.size();
            for(int i=0;i<levelSize;i++){
                TreeNode current = queue.pollFirst();
                if(current.left != null)
                    queue.add(current.left);
                if(current.right != null)
                    queue.add(current.right); 
            }
        }
        return maxDepth;
    }
}

Thinking 3: DFS

import javafx.util.Pair;
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;

        Stack<Pair<TreeNode,Integer>> stack =new Stack<>();
        stack.push(new Pair<>(root,1));

        int max =1;
        while(!stack.isEmpty()){
            Pair<TreeNode,Integer> current = stack.pop();

            Integer currentDepth=current.getValue();

            TreeNode currentNode=current.getKey();
            
            if(currentNode!= null){
                max=Math.max(currentDepth,max);
                stack.push(new Pair<>(currentNode.left,currentDepth+1));
                stack.push(new Pair<>(currentNode.right,currentDepth+1));
            }
        }
        return max;
    }
}