107. Binary Tree Level Order Traversal II

Problem:

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]
]

Analysis:
We can do normal level order and reverse the result. Another approach is DFS, create new List<Int> on the go.

Solution:

BFS:

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        Stack<List<Integer>> stack = new Stack<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            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);
            }
            stack.push(list);
        }

        while (!stack.isEmpty()) {
            res.add(stack.pop());
        }

        return res;
    }
}