【leetcode】Binary Tree Level Order Traversal I&II

II:

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode* root) {
vector<vector<int> >ret;
ret.clear();
if(root==NULL)
return ret;
queue<TreeNode*> Q;
Q.push(root);
Q.push(NULL);
vector<int> tmp;
while(!Q.empty())
{
TreeNode *p=Q.front();
Q.pop();
if(p!=NULL)
{
tmp.push_back(p->val);
if(p->left)
Q.push(p->left);
if(p->right)
Q.push(p->right);
}else{
if(!tmp.empty())
{
S.push(NULL);
ret.push_back(tmp);
tmp.clear();
}
}
}
reverse(ret.begin(),ret.end());
return ret;
}
};

I:

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]
]

class Solution {
public:
vector<vector<int> > levelOrder(TreeNode* root) {

queue<TreeNode *>Q;
vector<vector<int> > ret;
ret.clear();
vector<int> tmp;
Q.push(root);
Q.push(NULL);
TreeNode * node;

while(!Q.empty())
{
node=Q.front();

Q.pop();
if(node!=NULL)
{
tmp.push_back(node->val);
if(node->left)
Q.push(node->left);
if(node->right)
Q.push(node->right);
}else{
if(!tmp.empty())
{
Q.push(NULL);
ret.push_back(tmp);
tmp.clear();
}
}

}

return ret;
}
};