LeetCode – Refresh – Binary Tree Level Order Traversal ii

This question is almost same as the previous one.

I just use a stack (another vector) of vector<int> to store the level result. Then reverse it.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > result;
        if (!root) return result;
        stack<vector<int> > s;
        queue<TreeNode *> q;
        vector<int> level;
        q.push(root);
        int current = 1, future = 0;
        while (!q.empty()) {
            TreeNode *tmp = q.front();
            q.pop(), current--;
            if (tmp->left) {
                q.push(tmp->left);
                future++;
            }
            if (tmp->right) {
                q.push(tmp->right);
                future++;
            }
            level.push_back(tmp->val);
            if (!current) {
                current = future;
                future = 0;
                s.push(level);
                level.clear();
            }
        }
        while (!s.empty()) {
            result.push_back(s.top());
            s.pop();
        }
        return result;
    }
};