107 Binary Tree Level Order Traversal II

Approach 1 BFS

solve it as regular BFS then we reverse the output by res[::-1] or res.reverse(), as illustrated in the figure below

Code

from collections import deque
# 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]]:
        """
        top to bottom BFS, then UNO reverse card res[::-1]
        """
        if not root:
            return root

        queue = deque([root])
        levels = []

        while queue:
            level = []
            for _ in range(len(queue)):
                curr = queue.pop()
                level.append(curr.val)
                if curr.left:
                    queue.appendleft(curr.left)
                if curr.right:
                    queue.appendleft(curr.right)
            levels.append(level)
        return levels[::-1]