leetcode-102-Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:

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Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

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/**
* 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>> result;
if (!root) return result;
Helper(root, result, 0);
return result;
}
void Helper(TreeNode* root, vector<vector<int>> &result, int h){
if (!root) return;

if (result.size()<h+1){
vector<int> tmp;
tmp.push_back(root->val);
result.push_back(tmp);
}else{
result[h].push_back(root->val);
}
Helper(root->left, result, h+1);
Helper(root->right, result, h+1);
}
};
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();

if(root == null) return wrapList;

queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
List<Integer> subList = new LinkedList<Integer>();
for(int i=0; i<levelNum; i++) {
if(queue.peek().left != null) queue.offer(queue.peek().left);
if(queue.peek().right != null) queue.offer(queue.peek().right);
subList.add(queue.poll().val);
}
wrapList.add(subList);
}
return wrapList;
}
}
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