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:
Given binary tree [3,9,20,null,null,15,7],

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2
3
4
5
3
/
9 20
/
15 7

return its level order traversal as:

1
2
3
4
5
[
[3],
[9,20],
[15,7]
]

Solution

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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 {
public List<List<Integer>> levelOrder(TreeNode root) {
LinkedList<TreeNode> search = new LinkedList<TreeNode>();
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(root==null) return res;
search.add(root);
while(!search.isEmpty()){
int currentSize = search.size();
List<Integer> subList = new ArrayList<Integer>();
for(int i=0; i<currentSize;i++){
TreeNode node = search.poll();
subList.add(node.val);
if(node.left!=null) search.offer(node.left);
if(node.right!=null) search.offer(node.right);
}
res.add(subList);
}
return res;
}
}