Check if a given binary tree is complete (LaiCode, out-of-memory)

//class TreeNode { // public: // int value; // TreeNode* left; // TreeNode* right; // TreeNode(int v) : value(v), left(NULL), right(NULL) {} //}; class Solution { public: bool isCompleted(TreeNode* root) { // write your solution here if (root == nullptr || (root->left == nullptr && root->right == nullptr)) return true; queue< pair<TreeNode*, int> > q; TreeNode * new_node = root; // current node int new_level = 0; // initial level q.push(make_pair(new_node, new_level)); int curr_level = 0; int num_nodes_curr_level = 0; while(!q.empty()){ new_node = q.front().first; new_level = q.front().second; // operate on child nodes if (new_node->left != nullptr) q.push(make_pair(new_node->left, new_level+1)); if (new_node->right != nullptr) q.push(make_pair(new_node->right, new_level+1)); if (new_level == curr_level) num_nodes_curr_level++; else{ // we finished a level. Check if this level is complete. if (num_nodes_curr_level != pow(2, curr_level)) return false; else{ curr_level = new_level; num_nodes_curr_level++; } } } return true; } };