Boundary Traversal of binary tree

-
struct Node
{
    int data;
    Node *left, *right;
};

class Solution
{
public:
    void call_left(vector<int> &v, Node *root)
    {
        if (root == NULL)
            return;
        if (root->left == NULL && root->right == NULL)
            return;// don't push the leaf nodes.
        v.push_back(root->data);
        if (root->left)
            call_left(v, root->left);
        else
            call_left(v, root->right);
    }
    void call_leaves(vector<int> &v, Node *root)
    {
        if (root == NULL)
            return;
        if (root->left == NULL && root->right == NULL)
        {
            v.push_back(root->data);
        }
        call_leaves(v, root->left);
        call_leaves(v, root->right);
    }
    void call_right(stack<int> &s, Node *root)
    {
        if (root == NULL)
            return;
        if (root->left == NULL && root->right == NULL)
            return;// don't push the leaf nodes.
        s.push(root->data);
        if (root->right != NULL)
            call_right(s, root->right);
        else
            call_right(s, root->left);
    }
    vector<int> boundary(Node *root)
    {
        if (root == NULL)
            return {};
        vector<int> ans;
        if (root->left != NULL || root->right != NULL)
            ans.push_back(root->data); // for the root node

        call_left(ans, root->left); // for left nodes
        call_leaves(ans, root);     // for leaf nodes.
        stack<int> s; // for anticlockwise boundary
        // right nodes must be in reverse order, so use stack.
        call_right(s, root->right); // for right nodes.
        int temp;
        while (!s.empty())
        {
            // extract the right nodes from stack
            //  topmost right node will be at bottom.
            //  and the lowermost nodes will be at the
            //  bottom of the stack.
            //  Hence, while extracting the nodes,
            //  the nodes at lower level will come first.
            temp = s.top();
            s.pop();
            ans.push_back(temp);
        }
        return ans;
    }
};

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