102. Binary Tree Level Order Traversal (BFS) 20.3.15 Medium

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]
]

Solution:-------------------------------------------------- Queue & BFS ----- https://www.youtube.com/watch?v=B0n3gqPKKic
------------------------- Time complexity : O(n). Because we traverse the entire input tree once, the total run time is O(n), 
------------------------- where n is the total number of nodes in the tree.
------------------------- Space complexity : O(n)
------------------------- BFS will have to store at least an entire level of the tree in the queue (sample queue implementation). 
------------------------- With a perfect fully balanced binary tree, this would be (n/2 + 1) nodes (the very last level)
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        if root is None:
            return None
        queue, res = [root], []
        while queue:                                                                  
            currlist = []
            for i in range(len(queue)):
                pop_element = queue.pop(0)                                            # queue -> pop(0) not pop()!! First in First out!!
                if pop_element.left:
                    queue.append(pop_element.left)
                if pop_element.right: 
                    queue.append(pop_element.right)
                currlist.append(pop_element.val)
            res.append(currlist[:])
        return res                                                                    # Do not forget to initilize res in the beginning!
                                                                                      # res = []
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """


Java Version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }
        List<List<Integer>> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (q.size() != 0) {
            List<Integer> curr = new ArrayList<>();
            int currIterTime = q.size();
            for (int i = 0; i < currIterTime; i ++) {
                TreeNode popNode = q.poll();
                curr.add(popNode.val);
                if (popNode.left != null) {
                    q.offer(popNode.left);
                }
                if (popNode.right != null) {
                    q.offer(popNode.right);
                }
            }
            res.add(curr);
        }
        return res;
    }
}