Question

我打算像here一样进行二叉树搜索。我有一个问题是正确设置节点。

问题：当应该创建新节点时，根节点似乎被覆盖。第一次

Bintree.put(newValue)

被称为在Bintree.root中创建的新节点。第二次，根节点似乎在函数调用Bintree.put（newValue）中被覆盖。

以下这些行在执行时是否会更改根节点？

node = root
node = node.left # Left is a node below node
node = Node()

下面的行是我的程序的代码。

# Node for binary tree
class Node():
def __init__(self):
    self.data = None
    self.left = None
    self.right = None

class Bintree:
    def __init__(self):
            self.root = None

        def put(self, newvalue):
            '''Sends the new value into the correct position in the Bintree
            :param newvalue: The data that's sent to storage'''
            self.root = push(self.root, newvalue)

def push(root, value):
    # This function puts the value in the correct position in the bintree
    # Not to be used by user.
    node = root

    while node is not None:
        if node.data < value:
            node = node.left
        else:
            node = node.right

    node = Node()
    node.value = value

    return node

Answer 1

是的，你说得对，我把它拧了一下。

class Node():
    def init(self):
        self.data = None
        self.left = None
        self.right = None

class Bintree:
    def init(self):
        self.root = None

    def put(self, newvalue):
        '''Sends the new value into the correct position in the Bintree
        :param newvalue: The data that's sent to storage'''
        if self.root is None:
            self.root = Node()
            self.root.data = newvalue
        else:
            self.push(self.root, newvalue)

    def push(self, node, value):
        # This function puts the value in the correct position in the bintree
        # Not to be used by user.

        if value < node.data:
            if node.left is not None:
                self.push(node.left,value)
            else:
                node.left = Node()
                node.left.data = value
        else:
            if node.right is not None:
                self.push(node.right,value)
            else:
                node.right = Node()
                node.right.data = value

我是从头开始用递归做的。它更简单。当然它没有用，因为在你的第一次尝试中你总是将root设置为none，而在第二次你只是一直更新root（我的坏）

遍历链表python

1 个答案: