[LeetCode]Binary Tree Level Order Traversal II

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
//BFS is always work for the level order traverse of a tree
//all we need is add some additional information, to solve the specific problem of level order type
public:
vector<vector<int>> LevelOrder_BFS(TreeNode* root)
{
vector<vector<int>> ans;
if(!root)//put the null tree judge here will be more neet
return ans;
queue<pair<TreeNode*, int>> q;
int stepNow = 0;
q.push(make_pair(root, stepNow));
vector<int> curLevel;
while(!q.empty())
{
TreeNode* curNode = q.front().first;
int curStep = q.front().second;
q.pop();

if(curStep != stepNow)
{
ans.push_back(curLevel);
curLevel.clear();
stepNow = curStep;
}
curLevel.push_back(curNode->val);
if(curNode->left)
q.push(make_pair(curNode->left, curStep+1));
if(curNode->right)
q.push(make_pair(curNode->right, curStep+1));

}
ans.push_back(curLevel);//note: we should remember the last level here
reverse(ans.begin(), ans.end());
return ans;
}
vector<vector<int> > levelOrderBottom(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return LevelOrder_BFS(root);
}
};

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct Node
{
TreeNode* treeNode;
int level;
Node(TreeNode* _treeNode = NULL, int _level = 0):treeNode(_treeNode), level(_level){};
};
vector<vector<int> > levelOrderBottom(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return vector<vector<int> >();
queue<Node> nodeQ;
nodeQ.push(Node(root, 1));
int prevLevel = 0;
vector<int> curLevelAns;
vector<vector<int> > allLevelAns;
while(!nodeQ.empty())
{
TreeNode* curNode = nodeQ.front().treeNode;
int curLevel = nodeQ.front().level;
nodeQ.pop();
if(curNode == NULL) continue;
if(curLevel != prevLevel)
{
if(prevLevel != 0) allLevelAns.push_back(curLevelAns);
curLevelAns.clear();
}
prevLevel = curLevel;
curLevelAns.push_back(curNode->val);
nodeQ.push(Node(curNode->left, curLevel+1));
nodeQ.push(Node(curNode->right, curLevel+1));
}
allLevelAns.push_back(curLevelAns);
reverse(allLevelAns.begin(), allLevelAns.end());
return allLevelAns;
}
};