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] ]
思路：使用队列，等到这一层遍历结束再输出
void LevelOrder(BinTree* root)
{
if(root == NULL)
return ;
deque<BinTree*> de;
int num;
int curnum=0;
int i;
BinTree* node;
de.push_back(root);
vector<int> vec;
num =1;
while(!de.empty())
{
node = de.front();
de.pop_front();
num--;
vec.push_back(node->value);
if(node->left != NULL)
{
de.push_back(node->left);
curnum++;
}
if(node->right !=NULL)
{
de.push_back(node->right);
curnum++;
}
if(num ==0)
{
num =curnum;
curnum =0;
for(i=0;i<vec.size();i++)
cout<<vec[i]<<" ";
cout<<endl;
vec.clear();
}
}
} Binary Tree Level Order Traversal--LeetCode
原文：http://blog.csdn.net/yusiguyuan/article/details/44886293