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.

解题要点：

Preorder遍历树，最后检测小树是否是大树的一部分。

# 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
        """
        def preorder(t, left):
            if t == None:
                if left:
                    return "lNone"
                else:
                    return "rNone"
            return "#" + str(t.val) + " " + preorder(t.left, True) + " " + preorder(t.right, False)
        return preorder(t, True) in preorder(s, True)
        
        

找树的重合点，然后检测是否一样。

# 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
        """
        def identical(s, t):
            if s == None and t == None:
                return True
            if s == None or t == None:
                return False
            return s.val == t.val and identical(s.left, t.left) and identical(s.right, t.right)
        
        if t == None:
            return True
        if s == None:
            return False
        if identical(s, t):
            return True
        return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)