102. Binary Tree Level Order Traversal
Medium
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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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
这道题理解上有点难，花了一点时间想通。网上文章很多，但一开始还是感觉无法理解。因为网上讲细节的很少，贴代码的居多。我把自己的理解详细记录下来，哪天忘了，看看就明白了。
先学习了queue，queue是先进先出（FIFO）的。查了一下c++官网：www.cplusplus.com，里面的解释如下：
FIFO queue
queues are a type of container adaptor, specifically designed to operate in a FIFO context (first-in first-out), where elements are inserted into one end of the container and extracted from the other.
queues are implemented as containers adaptors, which are classes that use an encapsulated object of a specific container class as its underlying container, providing a specific set of member functions to access its elements. Elements are pushed into the "back" of the specific container and popped from its "front".
队列是一种容器适配器，专门设计用于在FIFO上下文（先进先出）中操作，其中元素插入容器的一端并从另一端提取。
队列作为容器适配器实现，这些类使用特定容器类的封装对象作为其基础容器，提供一组特定的成员函数来访问其元素。元素被推到特定容器的“后”中，并从其“前”弹出。
解题思路就是先把根节点push进queque这个容器，放入后，根节点就是这个queue的front。从front到end逐个处理queque中的treenode。拿根节点举例子，把根节点push进去后，queue容器中只有一个treenode，此时只需要把root的值取出来，压到最终要返回的vector中。压完后把root节点pop出来。这时queque中没有treenode了。然后我们来添加treenode，如果root->left有值，那么把root->left push到queque容器中，同样，如果root->right有值，那么把root->right push到queque容器中。然后重复，从front到end逐个处理queque中的treenode。知道queque容器中没有东西了。判断终止的条件就是queque.size()没有值了。
AC的代码如下：
/**
* 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>> y;
y.clear();
if(!root)return y;
queue<TreeNode*> que;
que.push(root);
while(que.size())
{
int num=que.size();
vector<int> x;
x.clear();
while(num>0)
{
TreeNode* t=que.front();
x.push_back(t->val);
que.pop();
if(t->left)que.push(t->left);
if(t->right)que.push(t->right);
num--;
}
y.push_back(x);
}
return y;
}
};