描述：
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]
]
方法：
做了這么多層次遍歷的變形了睡汹，終于做到原原本本的層次遍歷了酷愧。寫一下層次遍歷的方法吧不恭。使用queue數(shù)據(jù)結(jié)構(gòu)；設(shè)立last指針解阅，始終指向這一層的最后一個節(jié)點；設(shè)立nlast指針泌霍，始終記錄最新進入隊列的指針货抄。這樣每當(dāng)遍歷到last節(jié)點時，nlast就指向下一層最后的一個節(jié)點朱转。循環(huán)終止的條件時隊列中沒有元素了蟹地。
C++代碼：
/**
* 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> > levelOrder(TreeNode *root) {
vector<vector<int> > res;
if(!root)
return res;
vector<int> store;
queue<TreeNode*> qu;
qu.push(root);
TreeNode* last = root;
TreeNode* nlast;
while(qu.size())
{
TreeNode* tmp = qu.front();
qu.pop();
store.push_back(tmp->val);
if(tmp->left)
{
nlast = tmp->left;
qu.push(tmp->left);
}
if(tmp->right)
{
nlast = tmp->right;
qu.push(tmp->right);
}
if(tmp == last)
{
last = nlast;
res.push_back(store);
store.clear();
}
}
return res;
}
};
