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.

/** * 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) {        int level=getHeight(root);        vector<vector<int> > res(level);               levelOrder(res,root,level-1);        return res;    }    void levelOrder(vector<vector<int> > &res,TreeNode *root,int level){        if(root==NULL) return;        res[level].push_back(root->val);        if(root->left) levelOrder(res,root->left,level-1);        if(root->right) levelOrder(res,root->right,level-1);    }    int getHeight(TreeNode * root){        if(root==NULL) return 0;        int left=getHeight(root->left);        int right=getHeight(root->right);        return left>right?(left+1):(right+1);    }};