Leetcode: Binary Tree Level Order Traversal II

题目：

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]]

分析：

这道题目跟上道题目很相似 Leetcode: Binary Tree Level Order Traversal ，唯一不同的就是返回结果是从子叶节点到根节点，所以我们只需要将结果翻转下就好了！

参考代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution{private:    void visit(TreeNode *node, int depth, vector<vector<int>> &result)    {    if (node == NULL)    {    return;    }        if (result.size() > depth)    {    result[depth].push_back(node->val);    }    else    {    vector<int> level;    level.push_back(node->val);    result.push_back(level);    }        visit(node->left, depth + 1, result);    visit(node->right, depth + 1, result);    }public:    vector<vector<int>> levelOrderBottom(TreeNode *root)    {    vector<vector<int>> result;    visit(root, 0, result);        //就只有最后一句不一样，返回值对vector做了一个reverse    return vector<vector<int>>(result.rbegin(), result.rend());    }};