0199 Binary Tree Right Side View

Question

/**
 * 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 List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        from_right(root, res, 0);
        return res;
    }

    
    private void from_right(TreeNode root, List<Integer> res, int depth){
        if(root == null){
            return;
        }
        
        if(depth == res.size()){
            res.add(root.val);
        }
        
        from_right(root.right, res, depth+1);
        from_right(root.left, res, depth+1);
        
    }
}

Python

My Solution:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        q = collections.deque([root])

        while q:
            rightSide = None
            qLen = len(q)

            for i in range(qLen):
                node = q.popleft()
                if node:
                    rightSide = node
                    q.append(node.left)
                    q.append(node.right)
            if rightSide:
                res.append(rightSide.val)
                
        return res