层序遍历二叉树(Leetcode)

problem address

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

注意要找到当前层的所有节点继续向下遍历

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

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root: # NOTE: 注意root可能为空
            return []
        res = []
        cur_nodes = [root]
        next_nodes = []
        res.append([i.val for i in cur_nodes])
        while cur_nodes or next_nodes:
            for node in cur_nodes:
                if node.left:
                    next_nodes.append(node.left)
                if node.right:
                    next_nodes.append(node.right)
            if next_nodes:
                res.append(
                    [i.val for i in next_nodes]
                )

            cur_nodes = next_nodes
            next_nodes = []
        return res