题目链接：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}".

这道题的要求是从下往上分层遍历二叉树。

和Binary Tree Level Order Traversal同样的思路，只不过在最后需要将数组反转即可。

由于需要把每层的节点分别放入到数组中，因此需要引入变量n记录每层的节点数量。剩下的，就是广度优先搜索的方法了。

广度优先搜索算法（Breadth First Search），又叫宽度优先搜索，或横向优先搜索。从根节点开始，沿着树的宽度遍历树的节点。如果所有节点均被访问，则算法中止。借助队列数据结构，由于队列是先进先出的顺序，因此可以先将左子树入队，然后再将右子树入队。这样一来，左子树结点就存在队头，可以先被访问到。

时间复杂度：O(n)

空间复杂度：O(n)

 1 class Solution 2 { 3 public: 4     vector<vector<int> > levelOrderBottom(TreeNode *root) 5     { 6         vector<vector<int> > vvi; 7          8         if(NULL == root) 9             return vvi;10         11         queue<TreeNode *> q;12         q.push(root);13         while(!q.empty())14         {15             vector<int> vi;16             for(int i = 0, n = q.size(); i < n; ++ i)17             {18                 TreeNode *temp = q.front();19                 q.pop();20                 if(temp -> left != NULL)21                     q.push(temp -> left);22                 if(temp -> right != NULL)23                     q.push(temp -> right);24                 vi.push_back(temp -> val);25             }26             vvi.push_back(vi);27         }28         reverse(vvi.begin(), vvi.end());29         return vvi;30     }31 };