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 > res;
public:
void trval(TreeNode *root,int level)
{
if(root==NULL)
return;
if(level==res.size()) //这句是关键啊
{
vector v;
res.push_back(v);
}
res[level].push_back(root->val);
trval(root->left,level+);
trval(root->right,level+);
}
vector > levelOrder(TreeNode *root) { trval(root,);
return res; }
};

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]
]
/**
* 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<vector<int> > res;
void order(TreeNode *root,int level)
{
if(root==NULL)
return ; if(res.size()==level)
{
vector<int> tmp;
res.push_back(tmp);
} res[level].push_back(root->val);
order(root->left,level+);
order(root->right,level+);
}
vector<vector<int> > levelOrderBottom(TreeNode *root) {
order(root,);
return vector<vector<int> > (res.rbegin(),res.rend()); //直接一句话解决了 }
};

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