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] ]
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
許久未練, 生疏不少. 歎~
/**
* 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 > levelOrder(TreeNode *root)
{
vector vee;
vector>ve;
queue qu;
TreeNode *no;
if(NULL==root)
return ve;
qu.push(root);
int len = qu.size();
while(len!=0)
{
while(len--)
{
no = qu.front();
qu.pop();
vee.push_back(no->val);
if(NULL!=no->left)qu.push(no->left);
if(NULL!=no->right)qu.push(no->right);
}
ve.push_back(vee);
vee.clear();
len = qu.size();
}
return ve;
}
};