102. Binary Tree Level Order Tra

解法一. 使用队列.
根节点入队列，然后根节点出队列，输出根节点，其左右子节点分别入栈。
需要注意的是，此题要一层一层的输出子数组，注意控制一下循环条件就可以了

# 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]]:
        result = []
        if root is None:
            return result
        
        q = [root]
        while len(q):
            level = []
            length = len(q)
            for i in range(length):
                p = q.pop(0)
                level.append(p.val)
                if p.left:
                    q.append(p.left)
                if p.right:
                    q.append(p.right)
            result.append(level)
        return result
                
            
        
        

解法二 递归法

# 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]]:
        result = []
        if root is None:
            return result
        self.levelhelper(root, 0, result)
        return result
        
        
    def levelhelper(self, p, level, result):
        #当level等于数组的长度时，需要新申请一层用于存储
        if level == len(result):
            result.append([])
        if p:
            result[level].append(p.val)
        else:
            return
        if p.left:
            self.levelhelper(p.left, level+1, result)
        if p.right:
            self.levelhelper(p.right, level+1, result)