Binary Tree Level Order Traversal II

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

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

SOLUTION:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = [(root, 0)]
        levels = {}
        level = 0
        while len(queue) > 0:
            curr, level = queue.pop(0)
            levels[level] = levels.get(level, []) + [curr.val]
            if curr.left:
                queue.append((curr.left, level + 1))
            if curr.right:
                queue.append((curr.right, level + 1))
        return [levels[l] for l in range(level, -1, -1)]