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]
]

思路：

无非是最后使用reverse函数进行颠倒一下即可，不难。

代码：

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int> >res;
        if(root==NULL)  return res;
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty()){
            int n=q.size();
            vector<int>tem;
            for(int i=0;i<n;i++){
                TreeNode* temp=q.front();
                tem.push_back(temp->val);
                if(temp->left)   q.push(temp->left);
                if(temp->right)   q.push(temp->right);
                q.pop();
            }
            res.push_back(tem);
        }
        reverse(res.begin(),res.end());
        return res;
    }
};