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());
    }
};