题解 | 树的子结构

from types import SimpleNamespace
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pRoot1 TreeNode类 
# @param pRoot2 TreeNode类 
# @return bool布尔型
#
class Solution:
    def HasSubtree(self , pRoot1: TreeNode, pRoot2: TreeNode) -> bool:
        def issame(i,j):
            if not j :
                return True
            if not i or i.val != j.val:
                return False
            return issame(i.left,j.left) and issame(i.right,j.right) 
        def dfs(i,j):
            if not i or not j :
                return False
            return issame(i,j) or dfs(i.left,j) or dfs(i.right,j)
        return dfs(pRoot1,pRoot2)