Binary Tree Level Order Traversal II

Q:

 

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

[
  [15,7]
  [9,20],
  [3],
]

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

A:

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > results;
        if (!root) return results;
        vector<vector<TreeNode*> > nodes;
        nodes.push_back(vector<TreeNode*>());
        nodes.back().push_back(root);
        bool has_node = true;
        while (!nodes.back().empty()) {
            vector<TreeNode*> level_nodes = nodes.back();
            nodes.push_back(vector<TreeNode*>());
            for (int i = 0; i < level_nodes.size(); ++i) {
                if (level_nodes[i]->left) nodes.back().push_back(level_nodes[i]->left);
                if (level_nodes[i]->right) nodes.back().push_back(level_nodes[i]->right);
            }
        }
        for (int i = nodes.size() - 2; i >=0; --i) {
            results.push_back(vector<int>());
            for (int j = 0; j < nodes[i].size(); ++j) {
                results.back().push_back(nodes[i][j]->val);
            }
        }
        return results;
    }
};