Binary Tree Level Order Traversal II [LeetCode]

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],
]

BFS solution:

vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > ret;
if(root == NULL)
return ret;
vector<TreeNode *> level;
level.push_back(root);
while(true) {
if(level.size() == 0)
break;
vector<int> nums;
vector<TreeNode *> tmp;
for(auto item : level) {
nums.push_back(item->val);
if(item->left != NULL)
tmp.push_back(item->left);
if(item->right != NULL)
tmp.push_back(item->right);
}
ret.insert(ret.begin(), nums);
level = tmp;
}
return ret;
}

DFS solution:

void getLevelNums(TreeNode *root, vector<vector<int> > &ret, int level) {
if(ret.size() < level + 1){
vector<int> nums;
nums.push_back(root->val);
ret.insert(ret.begin(), nums);
}else if (ret.size() >= level + 1) {
ret[ret.size() - level - 1].push_back(root->val);
}

if(root->left != NULL)
getLevelNums(root->left, ret, level + 1);

if(root->right != NULL)
getLevelNums(root->right, ret, level + 1);

}
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > ret;
if(root == NULL)
return ret;
getLevelNums(root, ret, 0);
return ret;
}