LeetCode572. Subtree of Another Tree

最新推荐文章于 2025-02-03 15:37:10 发布

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LeetCode题目同时被 2 个专栏收录

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本文探讨了如何判断一棵二叉树是否为另一棵二叉树的子树，通过递归方法实现，并提供了详细的Python代码示例。

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.

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.

解题思路：

本题为判断t是否为s的子树，很明显，本题也是一个Divide And Conquer的题目，通过分治将原问题分成多个子问题来解决。原问题为判断以t为根节点的树是否为以s为根节点的树的子树。可分成三个子问题：以t为根节点的树即为以s为根节点的树；以t为根节点的树存在于以s的左儿子为根节点的树中；以t为根节点的树存在于以s的右儿子为根节点的树中。通过递归即可解决。

Python代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSubtree(self, s: 'TreeNode', t: 'TreeNode') -> 'bool':
        if s and not t:
            return False
        if not s and not t:
            return True
        if not s and t:
            return False
        if s.val == t.val:
            if self.isSametree(s, t):
                return True
        return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)
    
    def isSametree(self, s, t):
        if s and not t:
            return False
        if not s and not t:
            return True
        if not s and t:
            return False
        if s.val != t.val:
            return False
        else:
            return self.isSametree(s.left, t.left) and self.isSametree(s.right, t.right)
        return True