Binary Tree Level Order Traversal II

Q: 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],
]

按层（从上往下）输出二叉树的结点值。

A：vector insert方法的使用。每次访问到新的一层节点时，在result前部插入新的vector<int>.

 struct QueueNode{
     TreeNode *tree;
     int level;
     QueueNode(TreeNode *t,int l):tree(t),level(l){}
 };
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> result;
        if(!root)
            return result;
        
        queue<QueueNode> q;
        int levelcount = -1;
        q.push(QueueNode(root,0));
        while(!q.empty())
        {
            QueueNode qNode = q.front();
            q.pop();
            if(qNode.level!=levelcount)
            {
                vector<int> set;
                result.insert(result.begin(),set);
                levelcount++;
            }
            result[0].push_back(qNode.tree->val);
            if(qNode.tree->left)
                q.push(QueueNode(qNode.tree->left,qNode.level+1));
            if(qNode.tree->right)
                q.push(QueueNode(qNode.tree->right,qNode.level+1));
        }
        return result;
    }
};