十五天的时间,刷完了所有的简单题,避免遗忘,所以开始简单题的二刷,第一遍刷题的时候过得速度比较快,因为我觉得基础不好的我,不要硬着头皮去想最优的方法,而是应该尽量去学一些算法思想,所以每道题只给自己5-10分钟的时间想,想不出来的就去找相关的答案,所以刷的比较快。二刷的时候按照leetcode官方给出的题目分类展开,同时,将解题思路记录于加深印象。
想要一起刷题的小伙伴,我们一起加油吧!
我的github连接:https://github.com/princewen/leetcode_python
100. Same Tree
递归的判断两个数是否相等,并判断两棵树时候同时为空,即结构是否相等。
class Solution(object):
def isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
if p and q:
if p.val == q.val and self.isSameTree(p.left,q.left) and self.isSameTree(p.right,q.right):
return True
else:
return False
return p is q
104. Maximum Depth of Binary Tree
递归调用
# 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 maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
left = self.maxDepth(root.left)+1
right = self.maxDepth(root.right)+1
return max(left,right)
107. Binary Tree Level Order Traversal II
这里我们采用深度优先遍历的方式得到结果的输出:
(1) 树的根结点入栈
(2)判断栈是否为空,不为空,则出栈,并输出出栈树结点的值
(3)出栈树结点的右子树入栈
(4)出栈树结点的左子树入栈
(5)循环回到(2)
# 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 levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
stack = [(root,0)]
res = []
while stack!=[]:
node,level = stack.pop()
if node:
if len(res) < (level+1):
res.insert(0,[])
res[-(level+1)].append(node.val)
stack.append((node.right,level+1))
stack.append((node.left,level+1))
return res
108. Convert Sorted Array to Binary Search Tree
AVL tree 是一种特殊的二叉查找树,,首先我们要在树中引入平衡因子balance,表示结点右子树的高度减去左子树的高度差(右-左),对于一棵AVL树要么它是一棵空树,要么它是一棵高度平衡的二叉查找树,平衡因子balance绝对值不超过1.所以这里直接将一半元素放入左子树,一半元素放入右子树即可。
# 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 sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
if not nums:
return None
mid = len(nums) / 2
root = TreeNode(nums[mid])
root.left = self.sortedArrayToBST(nums[:mid])
root.right = self.sortedArrayToBST(nums[mid+1:])
return root
112. Path Sum
用递归的方法,只要有一条路径满足条件即可,而不是返回所有的可行路径,所以相对简单。
# 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 hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if not root:
return False
if root.left == None and root.right == None and root.val == sum:
return True
return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
226. Invert Binary Tree
就是将每个节点的左右子树互换,使用递归:
# 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 invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root:
root.left,root.right = self.invertTree(root.right),self.invertTree(root.left)
return root
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