Binary Tree Level Order Traversal， 二叉樹層級遍歷

[LeetCode] 

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

輸出如下：

[
  [15,7],
  [9,20],
  [3]
]

C++程序如下:

/**
 * Definition for binary tree
 * 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>> rvl;
        if(NULL == root) return rvl;
        queue<TreeNode*> qu;
        qu.push(root);
        while(!qu.empty()){
            vector<int> vec;
            int size = qu.size();
            for(int i = 0; i < size; i++){
                TreeNode *node = qu.front();
                qu.pop();
                vec.push_back(node->val);
                if(node->left != NULL) qu.push(node->left);
                if(node->right != NULL) qu.push(node->right);
             }
             rvl.push_back(vec);
        }
        //reverse
        int begin = 0; int end = rvl.size() - 1;
        while(begin < end){
            vector<int> tmp = rvl[begin];
            rvl[begin] = rvl[end];
            rvl[end] = tmp;
            begin++;
            end--;
        }
        return rvl;
    }
};