【面试题26】树的子结构

Python题解

# -*- 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):
        def Equal(num1, num2):
            if (num1 - num2 > -0.0000001) and (num1 - num2 < 0.0000001):return True
            else:return False
        def DoesTree1HasTree2(pRoot1, pRoot2):
            if not pRoot2:return True
            if not pRoot1:return False
            if not Equal(pRoot1.val, pRoot2.val):return False
            return DoesTree1HasTree2(pRoot1.left, pRoot2.left) and \
                    DoesTree1HasTree2(pRoot1.right, pRoot2.right)
        result = False
        if pRoot1 != None and pRoot2 != None:
            if Equal(pRoot1.val, pRoot2.val):
                result = DoesTree1HasTree2(pRoot1, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
        return result

考点

• 考查对二叉树遍历算法的理解及递归编程能力；
• 考查所写代码的鲁棒性，需要采用防御性编程的方式，每次访问指针地址都应考虑这个指针有没有可能是空。