作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址: https://leetcode.com/problems/complete-binary-tree-inserter/description/
题目描述:
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.
Write a data structure CBTInserter
that is initialized with a complete binary tree and supports the following operations:
CBTInserter(TreeNode root)
initializes the data structure on a given tree with head node root;CBTInserter.insert(int v)
will insert a TreeNode into the tree with value node.val = v so that the tree remains complete, and returns the value of the parent of the inserted TreeNode;CBTInserter.get_root()
will return the head node of the tree.
Example 1:
Input: inputs = ["CBTInserter","insert","get_root"], inputs = [[[1]],[2],[]]
Output: [null,1,[1,2]]
Example 2:
Input: inputs = ["CBTInserter","insert","insert","get_root"], inputs = [[[1,2,3,4,5,6]],[7],[8],[]]
Output: [null,3,4,[1,2,3,4,5,6,7,8]]
Note:
- The initial given tree is complete and contains between 1 and 1000 nodes.
- CBTInserter.insert is called at most 10000 times per test case.
- Every value of a given or inserted node is between 0 and 5000.
题目大意
编写一个完全二叉树的数据结构,需要完成构建、插入、获取root三个函数。函数的参数和返回值如题。
解题方法
周赛第三题,因为第二题我不会,就把这个题给放弃了……现在一看很简单啊。
完全二叉树是每一层都满的,因此找出要插入节点的父亲节点是很简单的。如果用数组tree保存着所有节点的层次遍历,那么新节点的父亲节点就是tree[(N -1)/2],N是未插入该节点前的树的元素个数。
构建树的时候使用层次遍历,也就是BFS把所有的节点放入到tree里。插入的时候直接计算出新节点的父亲节点。获取root就是数组中的第0个节点。
时间复杂度是O(N),空间复杂度是O(N)。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class CBTInserter(object):
def __init__(self, root):
"""
:type root: TreeNode
"""
self.tree = list()
queue = collections.deque()
queue.append(root)
while queue:
node = queue.popleft()
self.tree.append(node)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
def insert(self, v):
"""
:type v: int
:rtype: int
"""
_len = len(self.tree)
father = self.tree[(_len - 1) / 2]
node = TreeNode(v)
if not father.left:
father.left = node
else:
father.right = node
self.tree.append(node)
return father.val
def get_root(self):
"""
:rtype: TreeNode
"""
return self.tree[0]
# Your CBTInserter object will be instantiated and called as such:
# obj = CBTInserter(root)
# param_1 = obj.insert(v)
# param_2 = obj.get_root()
参考资料:
日期
2018 年 10 月 7 日 —— 假期最后一天!!