Substructure prove safety offer_ binary tree _

Sub-tree

Answers

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        result = False
        if pRoot1 and pRoot2:
            if pRoot1.val == pRoot2.val:
                result = self.isequal(pRoot1, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
        return result
    
    def isequal(self, root1, root2):
        if not root2:
            return True
        if not root1:
            return False
        if root1.val != root2.val:
            return False
        return self.isequal(root1.left, root2.left) and self.isequal(root1.right, root2.right)