【leetcode】107. Binary Tree Level Order Traversal II

bfs，利用两个队列，分别保存当前层和下一层的元素。

注意输出是从叶子到根节点，结点从左往右

/**
 *  @author         johnsondu
 *  @time           19:56 15th Oct 2015
 *  @status         Accepted
 *  @strategy       use bfs, two queue for store respect layer nodes
 *  @problem        Binary Tree Level Order Traversal II
 *  @url            https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
 */
/**
 * 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>> ans;
        if(root == NULL) return ans;
        queue<TreeNode*> q1;
        queue<TreeNode*> q2;
        TreeNode *cur = root;
        q1.push(cur);
        while(!q1.empty())
        {
            vector<int> tmp;
            while(!q1.empty()) {
                TreeNode* cur = q1.front();
                q1.pop();
                tmp.push_back(cur->val);
                if(cur->left != NULL) q2.push(cur->left);
                if(cur->right != NULL) q2.push(cur->right);
            }
            while(!q2.empty()) {
                q1.push(q2.front());
                q2.pop();
            }
            ans.push_back(tmp);
        }
        vector<vector<int>> result(ans.rbegin(), ans.rend());
        return result;
    }
};