题解 | #判断是不是完全二叉树# DFS递归 BFS队列
判断是不是完全二叉树
https://www.nowcoder.com/practice/8daa4dff9e36409abba2adbe413d6fae
- 深度优先用递归
- 广度优先用循环队列(一边出一边入)
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param root TreeNode类 # @return bool布尔型 # class Solution: def isCompleteTree(self , root: TreeNode) -> bool: if root is None: return True l = [root] # 是否遇到空节点 f = False while len(l) > 0: now = l.pop(0) # 遇到空节点就记录下 if now is None: f = True else: # 遇到之后还有不是空节点的就不是完全二叉树 if f: return False l.append(now.left) l.append(now.right) return True