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Commit 72d751d

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‎README.md‎

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@@ -73,7 +73,7 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
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|404|[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/#/description)| [Python [Yu]](./tree/Yu/404_sum_of_Left_Leaves.py) | _O(N)_| _O(1)_ | Easy ||[公瑾讲解](https://youtu.be/pH-KxPcBF_4)|
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|270|[Closest Binary Search Tree Value](https://leetcode.com/problems/closest-binary-search-tree-value/#/description)| [Python [Yu]](./tree/Yu/270.py)| _O(N)_| _O(1)_ | Easy ||[公瑾讲解](https://youtu.be/RwvQyKJxHZk)|
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|563|[Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/#/description)| [Python [Yu]](./tree/Yu/563.py) | _O(N)_| _O(1)_ | Easy | |[公瑾讲解](https://youtu.be/47FQVP4ynk0)|
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|538|[Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/#/description)|[Python [Yu]](./tree/Yu/538.py)|_O(N)_|_O(1)_| Easy ||[公瑾讲解](https://youtu.be/YoEPWtXmkD4)|
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## Tree Medium

‎math/453.py‎

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#!/usr/bin/python
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# -*- coding: utf-8 -*-
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# Author: Yu Zhou
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# 453. Minimum Moves to Equal Array Elements
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# 找到当前值和最小值的差,记录一个counter
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class Solution(object):
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def minMoves(self, nums):
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"""
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:type nums: List[int]
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:rtype: int
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"""
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count = 0
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#minu = min(nums)
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for num in nums:
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count += num - minu
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return count
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# 第一次的试验,TLE
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# 思路是每次进行一次sort,然后更改N-1的值
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class Solution(object):
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def minMoves(self, nums):
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"""
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:type nums: List[int]
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:rtype: int
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"""
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count = 0
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if len(nums) == 2:
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return max(nums) - min(nums)
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while not self.allEqual(nums):
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nums = sorted(nums)
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for num in xrange(len(nums)-1):
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num += 1
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count += 1
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return count
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def allEqual(self, nums):
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return len(set(nums)) <= 1

‎tree/Yu/109.py‎

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# Definition for singly-linked list.
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# class ListNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def sortedListToBST(self, head):
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"""
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:type head: ListNode
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:rtype: TreeNode
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"""
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# Edge
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if not head:
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return None
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if not head.next:
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return TreeNode(head.val)
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def findMid(head):
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slow, fast = head, head
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pre = None
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while fast and fast.next:
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pre = slow
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slow = slow.next
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fast = fast.next.next
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# right here the slow pointer is the mid point
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# ex: when list is 1,2,3,4. Slow pointer stops @ 3
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# ex2: when list is 1,2,3,4,5 Slow pointer stops @ 3 as well
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pre.next = None
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return slow
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mid = findMid(head)
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root = TreeNode(mid.val)
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root.left = self.sortedListToBST(head)
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root.right = self.sortedListToBST(mid.next)
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return root

‎tree/Yu/538.py‎

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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def convertBST(self, root):
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"""
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:type root: TreeNode
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:rtype: TreeNode
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"""
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# Input: The root of a Binary Search Tree like this:
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# 5
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# / \
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# 2 13
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self.sum = 0
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def dfs(root):
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if not root:
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return
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dfs(root.right)
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root.val += self.sum
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self.sum = root.val
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dfs(root.left)
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dfs(root)
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return root

‎tree/Yu/572.py‎

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#!/usr/bin/python
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# -*- coding: utf-8 -*-
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# Author: Yu Zhou
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# 572. Subtree of Another Tree
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# ****************
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# Descrption:
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# ****************
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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def isSubtree(self, s, t):
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:type t: TreeNode
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:rtype: bool
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"""
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#Edge
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if not s or not t:
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return False
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def isMatch(s, t):
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# 在每层的当前节点,做一套整体的比较
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# Edge
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if not s and not t:
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return True
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if not s or not t:
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return False
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if s.val == t.val:
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return isMatch(s.left, t.left) and isMatch(s.right, t.right)
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else:
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return False
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#Process
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if self.isMatch(s,t):
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if isMatch(s, t):
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return True
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#Recusion
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ifnots:
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returnFalse
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return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)
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def isMatch(self, s, t):
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if not s or not t:
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return s == t
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return s.val == t.val and self.isMatch(s.left, t.left) and self.isMatch(s.right, t.right)

‎tree/Yu/617.py‎

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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def mergeTrees(self, t1, t2):
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"""
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:type t1: TreeNode
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:type t2: TreeNode
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:rtype: TreeNode
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"""
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# PreOrder Append
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if not t1 and t2 or not t2 and t1:
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val1 = t1.val if t1 else 0
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val2 = t2.val if t2 else 0
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root = TreeNode(val1+val2)
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if not t1 and not t2:
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return root
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# DFS
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root.left = self.mergeTrees(t1.left, t2.left)
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root.right = self.mergeTrees(t1.right, t2.right)
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# Solved
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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def mergeTrees(self, t1, t2):
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"""
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:type t1: TreeNode
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:type t2: TreeNode
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:rtype: TreeNode
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"""
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if t2 and not t1:
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return t2
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if t1 and not t2:
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return t1
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if t1 and t2:
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root = TreeNode(t1.val + t2.val)
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root.left = self.mergeTrees(t1.left, t2.left)
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root.right = self.mergeTrees(t1.right, t2.right)
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return root

‎try.py‎

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