[September card ~ Leetcode daily question] 107. Level of the binary tree traversed II (difficulty: simple)

107. Level of the binary tree traversal II

A bifurcous tree is given back to the hierarchy of its node value from the bottom. (Ie, press the layer where the leaf node is located to the root node, from left to right from left to right)

Is this two days really weekend, a daily question of the barracks is simple.

This question is a BFS, the best time complexity is n, if you want to optimize, Python can use Deque, improve the speed when inserting and deleting (if the tree is relatively large), but the measurement speed is not much improved.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        ans = []
        bfs = collections.deque([root])

        if not root:
            return ans
            
        while(bfs):
            stand = []
            for _ in range(len(bfs)):    
                node = bfs.popleft()
                stand.append(node.val)
                if node.left:
                    bfs.append(node.left)
                if node.right:
                    bfs.append(node.right)
            ans.insert(0,stand)
        
        return ans

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