Binary Tree Level Order Traversal II.java

/*
Given a binary tree, return the bottom-up level order traversal of its nodes' values. 
(ie, from left to right, level by level from leaf to root).

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


time = O(n + n) = O(n)
space = O(n) for the queue and list
*/

/**
 * 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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                level.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                } 
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            result.add(level);
        }
        Collections.reverse(result);    // O(n)
        return result;
    }
}