LeetCode 102. Binary Tree Level Order Traversal

/**
* 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<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
vector<int> level;
if(!root) return res;

queue<TreeNode*> q1,q2;
q1.push(root);
while(!q1.empty()){
while(!q1.empty()){
TreeNode* cur = q1.front();
q1.pop();
level.push_back(cur->val);
if(cur->left) q2.push(cur->left);
if(cur->right) q2.push(cur->right);
}
res.push_back(level);
level.clear();
q1.swap(q2);

}
return res;
}
};

public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null)
return res;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
List<Integer> tmp = new ArrayList<Integer>();
int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode t = q.poll();
if (t.left != null)
q.offer(t.left);
if (t.right != null)
q.offer(t.right);
tmp.add(t.val);
}
res.add(tmp);
}
return res;
}
}

public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
dfs(res,root,0);
return res;
}
public void dfs(List<List<Integer>> list,TreeNode node,int deep){
if(node==null)return;
if(list.size()==deep)
list.add(new ArrayList<Integer>());
list.get(deep).add(node.val);
dfs(list, node.left, deep+1);
dfs(list, node.right, deep+1);
}
}