[LeetCode] 107. Binary Tree Level Order Traversal II

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]
]

Thought process:

We can do a regular level order traversal and reverse the result.

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 Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> levelOrder = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node == null) {
                    continue;
                }
                
                level.add(node.val);
                queue.offer(node.left);
                queue.offer(node.right);
            }
            
            if (level.size() > 0) {
                levelOrder.add(level);
            }
        }
        
        Collections.reverse(levelOrder);
        return levelOrder;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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