199. Binary Tree Right Side View

https://leetcode.com/problems/binary-tree-right-side-view/description/

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   
2     3         <---
      
  5     4       <---

You should return [1, 3, 4].

Sol 1:

recursion.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

/*
The core idea of this algorithm:

1.Each depth of the tree only select one node.

View depth is current size of result list.

*/

// recursion


public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        rightView(root, result, 0);
        return result;
    }
    
    public void rightView(TreeNode curr, List<Integer> result, int currDepth){
        if(curr == null){
            return;
        }
        if (currDepth == result.size()){
            result.add(curr.val);
        }
        
        rightView(curr.right, result, currDepth + 1);
        rightView(curr.left, result, currDepth + 1);
    }
}

Sol 2:

iteration

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */



// Time O(n) Space O(n)


public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (i == size - 1) {
                    // last element in current level
                    result.add(node.val);
                }
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
        }
        return result;
    }
}