Question

我花了好几个小时试图弄清楚如何做到这一点，我知道功能＆＃34; buildtree＆＃34;必须被递归地称为绘画，但我只是想弄清楚，我只需要一个用勺子喂的答案然后我可以研究，我已经尝试这么做了很长时间它不在我的能力范围内因为我无法理解逻辑事物的流动。（可悲的是我知道这只需要几行代码）

from ListBinaryTree import ListBinaryTree

def buildtree(left_inorder, left_preorder, right_inorder, right_preorder, root, tree):
    #dont know what to do.


def main(): 
    print("Binary Tree reconstructed by glee598:") 
    inorder_seq = input("Please enter the inorder sequence: ")
    preorder_seq = input("Please enter the preorder sequence: ")
    root = preorder_seq[0]
    root_inorder_index = inorder_seq.find(root)

    left_inorder = inorder_seq[:root_inorder_index]
    right_inorder = inorder_seq[root_inorder_index+1:]

    left_preorder = preorder_seq[1:preorder_seq.find(left_inorder[0])+1]
    right_preorder = preorder_seq[preorder_seq.find(left_inorder[0])+1:]

    tree = ListBinaryTree(root)

    buildtree(left_inorder, left_preorder, right_inorder, right_preorder, root, tree)
main()

ListBinaryTree类：

class ListBinaryTree:
        """A binary tree class with nodes as lists."""
        DATA = 0    # just some constants for readability
        LEFT = 1
        RIGHT = 2   

    def __init__(self, root_value, left=None, right=None):
        """Create a binary tree with a given root value
        left, right the left, right subtrees        
        """ 
        self.node = [root_value, left, right]

    def create_tree(self, a_list):
        return ListBinaryTree(a_list[0], a_list[1], a_list[2])

    def insert_value_left(self, value):
        """Inserts value to the left of this node.
        Pushes any existing left subtree down as the left child of the new node.
        """
        self.node[self.LEFT] = ListBinaryTree(value, self.node[self.LEFT], None)

    def insert_value_right(self, value):
        """Inserts value to the right of this node.
        Pushes any existing left subtree down as the left child of the new node.
        """      
        self.node[self.RIGHT] = ListBinaryTree(value, None, self.node[self.RIGHT])

    def insert_tree_left(self, tree):
        """Inserts new left subtree of current node"""
        self.node[self.LEFT] = tree

    def insert_tree_right(self, tree):
        """Inserts new left subtree of current node"""
        self.node[self.RIGHT] = tree

    def set_value(self, new_value):
        """Sets the value of the node."""
        self.node[self.DATA] = new_value

    def get_value(self):
        """Gets the value of the node."""
        return self.node[self.DATA]

    def get_left_subtree(self):
        """Gets the left subtree of the node."""
        return self.node[self.LEFT]

    def get_right_subtree(self):
        """Gets the right subtree of the node."""
        return self.node[self.RIGHT]

    def __str__(self):
        return '['+str(self.node[self.DATA])+', '+str(self.node[self.LEFT])+', '+\
 str(self.node[self.RIGHT])+']'

目标：

Answer 1

我认为你只是得到它，你只是在设置代码中放了太多的逻辑而没有意识到递归代码将是同一件事。

如果您更改递归buildtree函数的API，只需简单地获取整个inorder和preorder序列并返回由它们构建的树，就会轻松得多。< / p>

def buildtree(inorder, preorder):
    root_val = preorder[0]
    left_size = inorder.index(root_val) # size of the left subtree

    if left_size > 0:
        left = buildtree(inorder[:left_size], preorder[1:left_size+1])
    else:
        left = None

    if (left_size + 1) < len(inorder):
        right = buildtree(inorder[left_size+1:], preorder[left_size+1:])
    else:
        right = None

    return ListBinaryTree(root_val, left, right)

从中缀和前缀表达式重构二叉树[python]

1 个答案: