Leetcode 刷题Day7 404 SumofLeftLeaves 572 Subtree of Another Tree

自从昨天理解了python实现链表以后，今天理解python实现树就容易多了
树这种数据结构，理解递归思想对学习树很有帮助。




404 Sum of Left Leaves




Find the sum of all left leaves in a given binary tree.


Example:


    3
   / \
  9  20
    /  \
   15   7


There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.


解题思路：分类讨论
 按左child是叶子还是数讨论

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def sumOfLeftLeaves(self, root):        """        :type root: TreeNode        :rtype: int        """                        if not root: return 0        if root.left and not root.left.left and not root.left.right:#左子树不是树，是叶子            return root.left.val + self.sumOfLeftLeaves(root.right)        return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)#左子树不是叶子，是树

572. Subtree of Another Tree


Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.


Example 1:
Given tree s:


     3
    / \
   4   5
  / \
 1   2
Given tree t:
   4 
  / \
 1   2
Return true, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:


     3
    / \
   4   5
  / \
 1   2
    /
   0
Given tree t:
   4
  / \
 1   2
Return false.


解题思路：通过中左右这样的中序遍历把t树和s树转换为字符串，判断t字符串是否在s字符串里（借鉴自该题某discussion）

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def isSubtree(self, s, t):        """        :type s: TreeNode        :type t: TreeNode        :rtype: bool        def convert(p):            return "^" + str(p.val)  + convert(p.left) + convert(p.right) if p else "$"                return convert(t) in convert(s)

        

理解“^”在这里的目的，比如s是单节点树12,t是单节点数2，那么如果不加“^”，则会返回True,因为2在12里，但实际t并不在s里；所以加上“^”以后防止了这种情况，因为^2不在^12里，符合t不在s里的事实。
这种做法跑了115ms,打败了89%的用户。