二叉搜索树的底限和上限 Python

Question

我正在尝试计算 bst 的 ceil 和 floor 值，ceil 代码有效但我无法获得正确的 floor 值，我计算了 10 的 floor，在提供的测试用例中应该是10，但它 returns 9，我不确定为什么。

有什么想法吗？谢谢

class node():

    def __init__(self, key, data):

        self.data = data
        self.key = key
        self.leftnode = None
        self.rightnode = None


class binarysearch():

    def __init__(self):

        self.size = 0
        self.rootnode = None

    def insert(self, key, data):

        if self.rootnode == None:
            self.rootnode = node(key, data)
        else:
            self.insertnode(self.rootnode, key, data)

    def getroot(self):

        return self.rootnode

    def insertnode(self, root, key, data):

            if root.key == key:
                root.data = data

            elif key < root.key:
                if root.leftnode is None:                    
                    root.leftnode = node(key, data)
                else:
                    self.insertnode(root.leftnode, key, data)
            else:
                if root.rightnode is None:
                    root.rightnode = node(key, data)
                else:
                    self.insertnode(root.rightnode, key, data)

    def searchkey(self, key):

        if self.rootnode == None:
            return False
        else:
            found = self.searchnode(self.rootnode, key)
            if found:
                return True
            else:
                return False

    def searchnode(self, root, key):

        if root == None:
            return False

        else:
            if key == root.key:
                return True

            elif key < root.key:
                self.searchnode(root.leftnode, key)

            else:
                self.searchnode(root.rightnode, key)


    def floor(self, root, key): 

        if root == None:
            return None

        elif root.key == key:
            return key

        elif root.key > key:
            self.floor(root.leftnode, key)

        else:
            floorkey = self.floor(root.rightnode, key)
            if (floorkey is None) or (floorkey > key):
                return root.key
            else:
                return floorkey

    def ceil(self, root, key): 

        if root == None:
            return None

        elif root.key == key:
            return key

        elif root.key < key:
            return self.ceil(root.rightnode, key)

        else:             
            ceilkey = self.ceil(root.leftnode, key)
            if (ceilkey is None) or (ceilkey < key):
                return root.key
            else:
                return ceilkey



if __name__ == '__main__':

    a = binarysearch()
    a.insert(7,7)
    a.insert(1,1)
    a.insert(8,8)
    a.insert(3,3)
    a.insert(9,9)
    a.insert(2,2)
    a.insert(4,4)
    a.insert(11,11)
    a.insert(10,10)
    a.searchkey(7)
    root = a.getroot()
    ceil = a.ceil(root, 10)
    print(ceil)
    floor = a.floor(root, 10)
    print(floor)

Answer 1

您在 floor 函数的这一行缺少 return：

self.floor(root.leftnode, key)

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