26：树的子结构（剑指offer第2版Python）

1、题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

2、代码详解

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
# 递归终止条件：达到了树A或者树B的叶节点，即遍历完某树
class Solution:
    # 遍历函数
    def HasSubtree(self, pRoot1, pRoot2):
        # 空树不是任意树的子结构
        if pRoot1 == None or pRoot2 == None:
            return False

        # 如果val相等，才会遍历树2的左右节点，与树1 对比
        def hasEqual(pRoot1, pRoot2):
            if pRoot2 == None:  # 树2 已遍历完毕
                return True
            if pRoot1 == None:
                return False
            if pRoot1.val == pRoot2.val:
                if pRoot2.left == None:
                    leftEqual = True
                else:
                    leftEqual = hasEqual(pRoot1.left, pRoot2.left)
                if pRoot2.right == None:
                    rightEqual = True
                else:
                    rightEqual = hasEqual(pRoot1.right, pRoot2.right)
                # 左边和右边同时相等
                return leftEqual and rightEqual
            return False
        # 判断根节点是否相等
        if pRoot1.val == pRoot2.val:
            ret = hasEqual(pRoot1, pRoot2)
            if ret:
                return True
        # 判断左节点是否和pRoot2相等
        ret = self.HasSubtree(pRoot1.left, pRoot2)
        if ret:
            return True
        ret = self.HasSubtree(pRoot1.right, pRoot2)
        return ret

pRoot1 = TreeNode(8)
pRoot2 = TreeNode(8)
pRoot3 = TreeNode(7)
pRoot4 = TreeNode(9)
pRoot5 = TreeNode(2)
pRoot6 = TreeNode(4)
pRoot7 = TreeNode(7)
pRoot1.left = pRoot2
pRoot1.right = pRoot3
pRoot2.left = pRoot4
pRoot2.right = pRoot5
pRoot5.left = pRoot6
pRoot5.right = pRoot7

pRoot8 = TreeNode(8)
pRoot9 = TreeNode(9)
pRoot10 = TreeNode(2)
pRoot8.left = pRoot9
pRoot8.right = pRoot10

S = Solution()
print(S.HasSubtree(pRoot1, pRoot8))

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
# 递归终止条件：达到了树A或者树B的叶节点，即遍历完某树
class Solution:
    # 遍历函数
    def HasSubtree(self, pRoot1, pRoot2):
        # 设置标志位result，一旦匹配成功设为True，剩下的代码就不会执行
        # 如果匹配不成功，默认返回Flase
        result = False
        # 注意null的条件，HasSubTree中，如果两棵树都不为空才进行判断！！！
        if pRoot1 != None and pRoot2 != None:
            # 如果根节点值相同，递归调用函数
            # 查询是否具有相同的左、右子树结构
            if pRoot1.val == pRoot2.val:
                result = self.DoesTree1haveTree2(pRoot1, pRoot2)
            # 如果根节点不相同，则判断tree1的左子树和tree2是否相同，
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            # 再判断右子树和tree2是否相同
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
        return result
    # 判断函数
    def DoesTree1haveTree2(self, pRoot1, pRoot2):
        # 如果Tree2为空，则说明第二棵树遍历完了，即匹配成功
        if pRoot2 == None: #此处判断，包含两个情况：pRoot1为空/不为空，pRoot2为空，均True
            return True
        # tree1为空有两种情况
        # （1）如果tree1为空&&tree2不为空说明不匹配，
        #（2）如果tree1为空，tree2为空，说明匹配。
        if pRoot1 == None:
            return False

        if pRoot1.val != pRoot2.val:
            return False
        # 如果值相同，则递归地判断它们各自的左右节点的值是不是相同
        return self.DoesTree1haveTree2(pRoot1.left, pRoot2.left) and self.DoesTree1haveTree2(pRoot1.right, pRoot2.right)