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Problem: 102. 二叉树的层序遍历

思路

层序遍历，把同一层的节点的值放在同一个列表中

复杂度

时间复杂度: $O(n)$ n个节点

空间复杂度: $O(n)$ 列表和队列

Code

# 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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        def bfs(node):
            ls = []
            if not node: return ls
            
            q = collections.deque([node])
            while q:
                tem = [] # 存储每一层节点的值
                for _ in range(len(q)):
                    cur_node = q.popleft() # O(1) python list的pop(0)时间复杂度是O(n)
                    tem.append(cur_node.val)

                    if cur_node.left: q.append(cur_node.left)
                    if cur_node.right: q.append(cur_node.right)
                ls.append(tem)
            return ls
       
       
        return bfs(root)