二叉树的相关在线编程（python）

问题一：

　　输入一个整数数组，
　　判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。
　　假设输入的数组的任意两个数字都互不相同。

正确的后序遍历结果：

sequence = [37, 35, 51, 47, 59, 73, 93, 98, 87, 61]

python源码：

class Solution:
    def VerifySquenceOfBST(self, sequence):
        # write code here
        if sequence == None or len(sequence) == 0:
            return False
        length = len(sequence)
        root = sequence[length - 1]

        # 在二叉搜索 树中 左子树节点小于根节点   （应该是二叉排序树）
        for i in range(length):  # 获得左右的分叉口
            if sequence[i] > root:
                break

        # 二叉搜索树中右子树的节点都大于根节点
        for j in range(i, length):
            if sequence[j] < root:
                return False

        # 判断左子树是否为二叉树
        left = True
        if i > 0:
            left = self.VerifySquenceOfBST(sequence[0:i])

        # 判断 右子树是否为二叉树
        right = True
        if i < length - 1:
            right = self.VerifySquenceOfBST(sequence[i:-1])

        return left and right


if __name__ == '__main__':
    s = Solution()
    # sequence = [37, 35, 51, 47, 59, 73, 93, 98, 61, 87]   # 错误的二叉排序树的后续遍历
    sequence = [37, 35, 51, 47, 59, 73, 93, 98, 87, 61]     # 正确的二叉排序树的后续遍历
    print(s.VerifySquenceOfBST(sequence))

问题二： 

　　

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）

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:
    def HasSubtree(self, pRoot1, pRoot2):  # 判断是否有子树
        if pRoot1 == None or pRoot2 == None:
            return False

        result = False
        if pRoot1.val == pRoot2.val:  # 根节点相等，进行下一步
            result = self.IsSubtree(pRoot1, pRoot2) # 判断子节点是否“相等”

        if result == False:
            res1 = self.HasSubtree(pRoot1.left, pRoot2)
            res2 = self.HasSubtree(pRoot1.right, pRoot2)
            result = res1 or res2

        return result

    def IsSubtree(self, pRoot1, pRoot2):
        if pRoot2 == None:
            return True
        if pRoot1 == None:
            return False
        if pRoot1.val == pRoot2.val:
            leftt = self.IsSubtree(pRoot1.left, pRoot2.left)
            rightt = self.IsSubtree(pRoot1.right, pRoot2.right)
            return leftt and rightt
        return False

    def getBSTwithPreTin(self, pre, tin):  # 最终确定树的形状，使用object的形式存储
        if len(pre) == 0 | len(tin) == 0:  # 任何一个为空树，则返回空
            return None  # 直接跳出代码
        root = TreeNode(pre[0])  # 前序遍历直接找到根节点(第一个元素就是根节点)
        for i, item in enumerate(tin):  # 序号和元素，将中序遍历拆分成左右两支
            if root.val == item:
                root.left = self.getBSTwithPreTin(pre[1:i + 1], tin[:i])
                root.right = self.getBSTwithPreTin(pre[i + 1:], tin[i + 1:])
                return root


if __name__ == '__main__':
    solution = Solution()

    preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]  # 前序遍历
    middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]  # 中序遍历
    treeA = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)

    preorder_seq = [1, 2, 3]
    middleorder_seq = [2, 1, 3]
    treeB = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
    
    print(solution.HasSubtree(treeA, treeB))