C++ solution without a map

  • 0

    Instead of using a map for marking the vertical order, I just insert to the front of the result. This will have O(n) time for n level so the amortized time for insertion is still O(1). Therefore, the time complexity if O(num_nodes), space is O(num_nodes) because the BFS queue will have O(num_nodes) in the last level.

    vector<vector<int>> verticalOrder(TreeNode* root) {
        vector<vector<int>> result;
        if (!root)
            return result;
        queue<pair<TreeNode*, int>> bfsq;
        int level = 0, min_level = 0;
        bfsq.emplace(root, level);
        while (!bfsq.empty()) {
            auto cur = bfsq.front();
            int cur_level = cur.second;
            if (min_level > cur_level) {
                min_level = cur_level;
                result.insert(result.begin(), vector<int>());
            if (cur_level - min_level == result.size())
            result[cur_level - min_level].emplace_back(cur.first->val);
            if (cur.first->left)
                bfsq.emplace(cur.first->left, cur_level - 1);
            if (cur.first->right)
                bfsq.emplace(cur.first->right, cur_level + 1);
        return result;

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