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`@@ -137,7 +137,7 @@`
`137`	`137`	`\|40\|[Combination Sum II](https://leetcode.com/problems/combination-sum-ii/#/solutions)\| [Python ](./backtrack/Yu/39.py) \| _O(K * (2^N)_\| _O(N)_ \| [:tv:](https://youtu.be/HdS5dOaz-mk)\|`
`138`	`138`	`\|216\|[Combination Sum III](https://leetcode.com/problems/combination-sum-iii/#/description)\| [Python ](./backtrack/Yu/216.py) \| _O(K * (2^N)_\| _O(N)_ \|\|`
`139`	`139`	`\|17\|[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/description/)\| [Python ](./backtrack/Yu/17.py) \| _O(N*(4^N))_\| _O(N)_ \| [:tv:](https://www.youtube.com/watch?v=7KZWLM9QtRA)\|`
`140`		`-`
	`140`	`+\|131\|[Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning/description/)\|[Python ](./backtrack/Yu/131.py)\|_O(N*(2^N))_\|_O(N)_\|[:tv:](https://youtu.be/UFdSC_ml4TQ)\|`
`141`	`141`
`142`	`142`	`## Greedy Medium`
`143`	`143`	`\| # \| Title \| Solution \| Time \| Space \| Video\|`

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`@@ -12,23 +12,18 @@`
`12`	`12`	`# 思路:`
`13`	`13`	`# 这道题只要知道一个规律,就是左边Child要等于右边Child,就很容易了`
`14`	`14`	`# 先解决Root的Edge,之后在对其他进行一个统一处理,我选择写一个Helper,也可以不写`
`15`		`-# ****************`
`16`		`-# Final Solution *`
`17`		`-# Recursive *`
`18`		`-# ****************`
	`15`	`+`
	`16`	`+# if not left or not right: return left == right的意思是:`
	`17`	`+ # if not left or not right: return False`
	`18`	`+ # if not left and not right: return True`
	`19`	`+`
`19`	`20`	`class Solution(object):`
`20`	`21`	`def isSymmetric(self, root):`
`21`		`- """`
`22`		`- :type root: TreeNode`
`23`		`- :rtype: bool`
`24`		`- """`
`25`		`- if not root:`
`26`		`- return True`
`27`		`- else:`
`28`		`- return self.isEqual(root.left, root.right)`
`29`	`22`
`30`		`- def isEqual(self, n1, n2):`
`31`		`- if not n1 or not n2:`
`32`		`- return n1 == n2`
`33`		`- else:`
`34`		`- return n1.val == n2.val and self.isEqual(n1.left, n2.right) and self.isEqual(n1.right, n2.left)`
	`23`	`+ if not root: return True`
	`24`	`+ def dfs(left, right):`
	`25`	`+ if not left or not right: return left == right`
	`26`	`+ if left.val != right.val: return False`
	`27`	`+ return dfs(left.left, right.right) and dfs(left.right, right.left)`
	`28`	`+`
	`29`	`+ return dfs(root.left, root.right)`

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