【LectCode】102. Binary Tree Level Order Traversal

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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

解题思路：把树的节点依次加入队列中，用一个整数count来计算当前层节点的数目，用一个while循环则可以把当前层的节点的值读入向量中，并在访问当前层节点的时候将其后继加入队列中，访问完毕后，删除节点。

解答：

/**
 * 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> > vec;
        if(root == NULL){
        return vec;
} 
else{
queue<TreeNode*> q;
q.push(root);
int count = 0;
TreeNode* node;
while(!q.empty()){
count = q.size();
vector<int> a;
while(count){
node = q.front();
a.push_back(node->val);
if(node->left != NULL){
q.push(node->left); 
}
if(node->right != NULL){
q.push(node->right); 
}
q.pop();
count --;
}
vec.push_back(a); 
a.clear();
}
}
return vec;
    }
};