572. Subtree of Another Tree

最新推荐文章于 2024-09-25 18:14:06 发布

原创最新推荐文章于 2024-09-25 18:14:06 发布 · 187 阅读

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本文介绍了一种通过递归方法来判断一棵二叉树是否为另一棵二叉树的子结构的有效算法，并提供了一个创新性的解决方案，即将二叉树转换为字符串进行比较。

题目描述：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.

示例：

Example 1:
Given tree s:

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:

Given tree t:

   4
  / \
 1   2

Return false.

思想：树的题目都是递归来的多，首先应该判断相同部分是否从根节点开始，若不是递归判断左右子树。

class Solution:
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        if s is None or t is None:
            return s is None and t is None
        if self.isSame(s, t):
            return True
        else:
            return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)
            
    def isSame(self, s, t):
        if not s:
            return not t
        if not t:
            return not s
        left = self.isSame(s.left, t.left)
        mid = (s.val == t.val)
        right = self.isSame(s.right, t.right)
        return left and mid and right

展示一位大神的代码，把树打印成字符串，再比较字符串t是否在s中，这种做法期初会有个漏洞，就是第二种例子的情况，本做法用l,r分别标识就巧妙地避免了这种尴尬。太赞了

class Solution:
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        def treeToString(root, s = ""):
            if root:
                return treeToString(root.left,"l") + str(root.val) + treeToString(root.right, "r")
            return s
        
        stringS = treeToString(s)
        stringT = treeToString(t)
        
        if stringT not in stringS:
            return False
        
        return True