Binary Tree Level Order Traversal II

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

Solution:

/** * 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:    void dfs(vector<vector<int> > &res, TreeNode* root, int depth)    {        if(root == NULL) return;        if(res.size() > depth)        {            res[depth].push_back(root->val);        }        else        {            vector<int> v;            v.push_back(root->val);            res.push_back(v);        }        dfs(res, root->left, depth + 1);        dfs(res, root->right, depth + 1);    }    vector<vector<int>> levelOrderBottom(TreeNode* root) {        vector<vector<int> > res;        dfs(res, root, 0);        reverse(res.begin(), res.end());        return res;    }};