【LeetCode】958. Check Completeness of a Binary Tree 解題報告(Python)
阿新 • • 發佈:2018-12-30
作者: 負雪明燭
id: fuxuemingzhu
個人部落格: http://fuxuemingzhu.cn/
目錄
題目地址:https://leetcode.com/problems/check-completeness-of-a-binary-tree/
題目描述
Given a binary tree, determine if it is a complete binary tree.
Definition of a complete binary tree from Wikipedia:
- In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Example 1:
Input: [1,2,3,4,5,6] Output: true Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.
Example 2:
Input: [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn't as far left as possible.
Note:
- The tree will have between 1 and 100 nodes.
題目大意
判斷一個二叉樹是不是完全二叉樹。
解題方法
BFS
這個題可以使用DFS或者BFS先找出二叉樹的層次遍歷。之後的判斷中,使用DFS比較麻煩一些。
使用BFS的話層次遍歷比較簡單,因為我們從每層的從左到右進行遍歷,如果某一層已經出現None之後,後面還有非空葉子節點的話,那麼就不是完全二叉樹。
程式碼如下:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isCompleteTree(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root: return True
res = []
que = collections.deque()
que.append(root)
hasNone = False
while que:
size = len(que)
for i in range(size):
node = que.popleft()
if not node:
hasNone = True
continue
if hasNone:
return False
que.append(node.left)
que.append(node.right)
return True
DFS
思路是,除了最後一層之外,其餘的層必須都是滿二叉樹,最後一層左邊只能全部是非空葉子節點,如果出現None之後,後面不能再有非空葉子節點了。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isCompleteTree(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root: return True
res = []
self.getlevel(res, 0, root)
depth = len(res) - 1
for d in range(depth):
if d != depth - 1:
if None in res[d] or len(res[d]) != (2 ** d):
return False
else:
ni = -1
for i, n in enumerate(res[d]):
if n == None:
if ni == -1:
ni = i
else:
if ni != -1:
return False
return True
def getlevel(self, res, level, root):
if level >= len(res):
res.append([])
if not root:
res[level].append(None)
else:
res[level].append(root.val)
self.getlevel(res, level + 1, root.left)
self.getlevel(res, level + 1, root.right)
日期
2018 年 12 月 16 日 —— 周賽好難