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.

Solution 1: recursive

我们先从s的根结点开始，跟t比较，如果两棵树完全相同，那么返回true，否则就分别对s的左子结点和右子结点调用递归再次来判断是否相同，只要有一个返回true了，就表示可以找得到。

Time complexity: O(m*n). m is the number of nodes in s, n is the number of nodes in t.

Space complexity: 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 Solution(object):
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        if s == None:
            return False
        if self.isSame(s, t):
            return True
        return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)

    def isSame(self, s, t):
        if s == None and t == None:
            return True
        if s == None or t == None:
            return False
        if s.val != t.val:
            return False
        return self.isSame(s.left, t.left) and self.isSame(s.right, t.right)

Solution 2: serialize tree and compare

下面这道题的解法用到了之前那道 Serialize and Deserialize Binary Tree 的解法，思路是对s和t两棵树分别进行序列化，各生成一个字符串，如果t的字符串是s的子串的话，就说明t是s的子树，但是需要注意的是，为了避免出现[12], [2], 这种情况，虽然2也是12的子串，但是[2]却不是[12]的子树，所以我们再序列化的时候要特殊处理一下，就是在每个结点值前面都加上一个字符，比如 在前面加‘^’ 后面加‘#’ 来分隔开，那么[12]序列化后就是"^12#"，而[2]序列化之后就是"^2#"，这样就可以完美的解决之前的问题了。最低端的空leaf 赋值 ‘$’.

一定要用Preorder traverase. 因为要保证根点下面的所有点都相同。

class Solution(object):
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        def preorder(node):
            if not node:
                return "$"
            return "^" + str(node.val) + "#" + preorder(node.left) + preorder(node.right)
        return preorder(t) in preorder(s)