Leetcode Binary Tree Right Side View problem solution

In this Leetcode Binary Tree Right Side View problem solution we have Given the root of 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.

Problem solution in Python.

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

Problem solution in Java.

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;
    }
}

Problem solution in C++.

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);
    }
};

Problem solution in C.

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;
}

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