判断完全二叉树

# 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

# 判断一个树是否是完全二叉树
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解决思路：
1、任一节点，有右孩子无左孩子，return False
2、在1不违规的条件下，如果遇到了第一个左右孩子不全的节点，后续节点均为叶子节点
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from collections import deque
def is_valid_cbt(head: TreeNode) -> bool:
    # 采用宽度优先遍历，需使用队列作为辅助
    if head is None:
        return True
    queue = deque()
    # 是否遇到过左右两个孩子节点不双全的节点
    leaf = False
    queue.append(head)
    while queue:
        head = queue.popleft()
        l = head.left
        r = head.right
        # 条件2
        if (leaf and (l is not None or r is not None)) \
            # 条件1
            or (l is None and r is not None):
            return False

        if l is not None:
            queue.append(l)
        if r is not None:
            queue.append(r)
        if l is None or r is None:
            leaf = True
    return True