【Binary Tree Level Order Traversal】cpp

题目：

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

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

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>> levelOrder(TreeNode* root) {
            vector<vector<int> > ret;
            if (!root) return ret;
            vector<int> tmp_ret;
            deque<TreeNode *> currLevel, nextLevel;
            currLevel.push_back(root);
            while ( !currLevel.empty() )
            {
                while ( !currLevel.empty() )
                {
                    TreeNode * tmp = currLevel.front();
                    currLevel.pop_front();
                    tmp_ret.push_back(tmp->val);
                    if ( tmp->left ) nextLevel.push_back(tmp->left);
                    if ( tmp->right ) nextLevel.push_back(tmp->right);
                }
                ret.push_back(tmp_ret);
                tmp_ret.clear();
                std::swap(currLevel, nextLevel);
            }
            return ret;
    }
};

tips：

核心：两个队列技巧

1. 采用两个队列，一个队列存放本层TreeNode，另一个队列存放下一层的TreeNode

2. 本层Node逐个出队的同时加入tmp_ret的vector，下一层的Node逐个入队

3. 本层Node全部出队之后，tmp_ret推入ret（并注意清空tmp_ret，第一次没有清空tmp_ret没有AC）

4. 逐个时候currLevel队列已经空了，nextLevel队列存放的都是下一层的节点，利用swap操作交换二者。（这个交换是O(1)的指针交换）

完毕。

======================================

一些关于二叉树 deque queue的参考资料：

http://www.cnblogs.com/way_testlife/archive/2010/10/07/1845264.html

http://stackoverflow.com/questions/2247982/deque-vs-queue-in-c

=======================================

第二次过这道题，用双队列的思路实现的，代码一次AC。

/**
 * 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>> levelOrder(TreeNode* root) {
            vector<vector<int> > ret;
            queue<TreeNode*> curr;
            queue<TreeNode*> next;
            if ( root ) curr.push(root);
            while ( !curr.empty() )
            {
                vector<int> tmp;
                while ( !curr.empty() )
                {
                    tmp.push_back(curr.front()->val);
                    if ( curr.front()->left ) next.push(curr.front()->left);
                    if ( curr.front()->right ) next.push(curr.front()->right);
                    curr.pop();
                }
                ret.push_back(tmp);
                std::swap(next, curr);
            }
            return ret;
    }
};