Leetcode Binary Tree Right Side View problem solution

class Solution:
    def rightSideView(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        q = [root]
        res = []
        while q:
            n = len(q)
            for i in range(n):
                popped = q.pop(0)
                if popped.left:
                    q.append(popped.left)
                if popped.right:
                    q.append(popped.right)
                if i == n - 1:
                    res.append(popped.val)
        return res

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if(root==null) return result;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        while(q.size()>0){
            int size = q.size();
            for(int i=0;i<size;i++){
                TreeNode node= q.poll();
                if(i==size-1)
                    result.add(node.val);
                if(node.left!=null) q.add(node.left);
                if(node.right!=null) q.add(node.right);
            }
        }
        return result;
    }
}

class Solution {
    int mxl=-1;
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ans;
        dfs(root,ans,0);
        return ans;
    }
    
    void dfs(TreeNode* root, vector<int>& ans,int level)
    {
        if(root==nullptr) return;
        if(mxl<level)
        {
            ans.push_back(root->val);
            mxl = level;
        }
        dfs(root->right,ans,level+1);
        dfs(root->left,ans,level+1);
    }
};

int* rightSideView(struct TreeNode* root, int* returnSize){
    struct TreeNode *n[128] = { root };
    int *a = malloc(sizeof(int [100])), k = 0;
    for (int f = 0, b = 1, lb ; root && f < b ; a[k++] = n[lb - 1]->val)
        for (lb = b ; f < lb && b < 128 ; (n[b++] = n[f]->left) || b--, (n[b++] = n[f++]->right) || b--);
    return *returnSize = k, a;
}