102 Binary Tree Level Order Traversal

102 Binary Tree Level Order Traversal


class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
      List<List<Integer>> result = new ArrayList<>();
      if(root == null) return result;
      Queue<TreeNode> queue = new LinkedList<>(); // queue, stack interface??
      queue.offer(root);
      while(!queue.isEmpty()){
        int size = queue.size();
        List<Integer> list = new ArrayList<>(); // create a new list here before the size 
        for(int i = 0; i < size; i++){
          TreeNode cur = queue.poll();
          list.add(cur.val);
          if(cur.left != null) queue.offer(cur.left);
          if(cur.right != null) queue.offer(cur.right);
        }
        result.add(list);
    }
    return result;
    }
}


// stack , push , pop
// queue. Offer, poll

 

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],

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]