102. 二叉树的层序遍历

https://leetcode.cn/problems/binary-tree-level-order-traversal/

• 提交时间:2023-02-21 12:30:54
• 执行用时：48 ms, 在所有 Python3 提交中击败了10.85%的用户
• 内存消耗：15.8 MB, 在所有 Python3 提交中击败了11.27%的用户
• 通过测试用例：34 / 34

# 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: TreeNode) -> List[List[int]]:
        retList = []
        self.getNode(0, root, retList)
        return retList
    
    def getNode(self, index, child: TreeNode, retList):
        if None == child:
            return
        if len(retList) < index + 1:
            retList.append([child.val])
        else:
            retList[index].append(child.val)
        self.getNode(index + 1, child.left, retList)
        self.getNode(index + 1, child.right, retList)