How to Sum the Root To Leaf in Binary Numbers in a Binary Tree using Breadth First Search?

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumRootToLeaf(TreeNode* root) {
        int sum = 0;
        queue<pair<TreeNode*, int>> q;
        if (root == NULL) return 0;
        q.push(make_pair(root, root->val)); // push the root
        while (q.size() > 0) {
            auto p = q.front();
            q.pop();
            TreeNode* node = p.first;
            int cur = p.second;
            if (node->left == NULL && node->right == NULL) { // leaf node
                sum += cur;   
            } else {
               if (node->left != NULL) { // push left children
                   q.push(make_pair(node->left, cur * 2 + node->left->val));
               }
               if (node->right != NULL) { // push right children
                   q.push(make_pair(node->right, cur * 2 + node->right->val));
               }
            }
        }
        return sum;
    }
};

import javafx.util.Pair;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumRootToLeaf(TreeNode root) {
        int sum = 0;
        if (root == null) return 0;
        Queue<Pair<TreeNode, Integer>> q = new LinkedList<Pair<TreeNode, Integer>>();
        q.add(new Pair(root, root.val));
        while (!q.isEmpty()) {
            Pair<TreeNode, Integer> p = q.poll(); // pop and return the node from the queue
            TreeNode node = p.getKey();
            int cur = p.getValue();
            if (node.left == null && node.right == null) { // sum up the from root to current leaf node
                sum += cur;
            } else {
                if (node.left != null) { // enqueue the left child
                    q.add(new Pair(node.left, cur * 2 + node.left.val));
                }
                if (node.right != null) { // enqueue the right child
                    q.add(new Pair(node.right, cur * 2 + node.right.val));
                }                
            }            
        }
        return sum;
    }
}