Level Order

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

Example :
Given binary tree

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Also think about a version of the question where you are asked to do a level order traversal of the tree when depth of the tree is much greater than number of nodes on a level.

Solution:

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode A) {
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(A);
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i ++) {
                TreeNode curr = queue.poll();
                level.add(curr.val);
                if (curr.left != null) {
                    queue.offer(curr.left);
                }
                if (curr.right != null) {
                    queue.offer(curr.right);
                }
            }
            result.add(level);
        }
        return result;
    }
}