leetcode---Binary Tree Right Side View---层次遍历

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:

Given the following binary tree,

1            <---
/   \
2     3         <---
\     \
5     4       <---

You should return 

[1, 3, 4]

.

/**
* 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<int> ans;
vector<int> rightSideView(TreeNode* root)
{
queue<TreeNode*> q;
if(root)
q.push(root);
q.push(NULL);
while(!q.empty())
{
TreeNode *node = q.front();
q.pop();
if(node == NULL)
{
if(q.empty())
break;
else
q.push(NULL);
}
else
{
if(q.front() == NULL)
ans.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}
}
return ans;
}
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def rightSideView(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
queue = []
List = []
if(root != None):
queue.append(root)
queue.append(None)
while(queue):
node = queue.pop(0)
if(node == None):
print 1
if(not queue):
break
else:
queue.append(None)
else:
print node.val
if(queue[0] == None):
List.append(node.val)
if(node.left != None):
queue.append(node.left)
if(node.right != None):
queue.append(node.right)
return List