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]]

思路：按层序遍历得到结果，然后逆序输出就ok了。

class Solution {public:    vector<vector<int> > levelOrderBottom(TreeNode *root) {        vector<vector<int> >res;        res.clear();        if(root == NULL) {            return res;        }        queue<TreeNode*> Q1;        Q1.push(root);        vector<int> a;        while(!Q1.empty()) {            queue<TreeNode*> Q2;            a.clear();            while(!Q1.empty()) {                TreeNode* top = Q1.front();                Q1.pop();                if (top->left != NULL) {                    Q2.push(top->left);                }                if (top->right != NULL) {                    Q2.push(top->right);                }                a.push_back(top->val);            }            Q1 = Q2;            res.push_back(a);        }        vector<vector<int> > rRes;        int len = res.size();        int i;        for(i=len-1; i>=0; i--) {            rRes.push_back(res[i]);        }            return rRes;    }};