二叉树遍历功能回复步骤,因此不能多次运行它?

时间:2016-03-28 11:39:28

标签: python algorithm binary-tree recursive-datastructures

我实现了二叉树有序遍历功能。基本上有3个递归步骤:去左边的孩子,获取货物数据,去找右边的孩子。所以我设计了一个增量标志(一个属于Node类的属性)来记录在遍历期间是否已经采取特定节点上的步骤。如果我运行一次,标志运行良好。当第二次运行它时,标志违背了目的。

解决方案:我可以使用与我用来生成节点对象以重置标志的功能类似的功能。但它似乎非常多余并重复我的自我。您是否可以为我提供更好的解决方案来重置遍历目的的标志或不使用这些步骤的不同解决方案?

谢谢! 下面是Python中的实现:

"""implementation of Binary Tree"""


class BinaryTreeNode(object):

    def __init__(self, data, left=None, right=None, parent=None):
        self.data = data
        self.left = left
        self.right = right
        self.parent = parent
        self.traversal_step = int(0)

    def __str__(self):
        return str(self.data)

    def get_data(self):
        return self.data

    def get_left(self):
        return self.left

    def get_right(self):
        return self.right

    def get_parent(self):
        return self.parent

    def set_left(self, left):
        self.left = left

    def set_right(self, right):
        self.right = right

    def set_parent(self, parent):
        self.parent = parent

    def set_traversal_step(self, reset=False):
        if reset == False:
            self.traversal_step += 1

        else:
            self.traversal_step = 0

    def get_traversal_step(self):
        return self.traversal_step


class BinaryTree(object):
    """implement a binary tree
    Protocol:
    any data has value less than value of its parent node
    will be placed on the left child node. While the ones
    greater, will be placed to the right child node
    """
    def __init__(self):
        self.root = None
        self.tree_depth = int(0)
        self.node_sum = int(0)

    def insert(self, data):
        new_node = BinaryTreeNode(data)
        current_node = self.root
        # print('begin inserting : ' + str(data))
        if self.root:
            # Determine left/right side should be chosen for the new node
            fulfill_status = False
            while not fulfill_status:
                if data >= current_node.get_data():

                    if current_node.get_right():
                          # print('move to RIGHT, and dive to next level')
                        current_node = current_node.get_right()
                    else:
                        current_node.right = new_node
                        new_node.set_parent(current_node)
                        fulfill_status = True
                else:
                    if current_node.get_left():
                          # print('move to LEFT, and dive to next level')
                        current_node = current_node.get_left()
                    else:  # empty node slot found
                        current_node.left = new_node
                        new_node.set_parent(current_node)
                        fulfill_status = True
                # 3. verify status on the current node
                  # print('Current parent node = ' + str(current_node.get_data()))
                  # print('Child status: '
                  #     + 'left=' + str(current_node.get_left())
                  #     + ' right=' + str(current_node.get_right()))
                  # print('new child\'s parent node is:' + str(new_node.get_parent()))

        else:
            # print('Building a new tree now, root = ' + str(data))
            self.root = new_node

        # print('Finishing inserting...' + '#' * 30)

    def query(self, data):
        """check if the data presents in the Tree already"""
        current_node = self.root
        print('begin querying data : {} '.format(data) + '#' * 50)
        if self.root:
            # Determine left/right side should be chosen for the new node
            found_status = False
            while not found_status:
                if data == current_node.get_data():
                    found_status = True
                    break
                elif data > current_node.get_data():
                    if current_node.get_right():
                        # print('move to RIGHT, and dive to next level')
                        current_node = current_node.get_right()
                    else:
                        break  # no existing node larger than the current node.
                else:
                    if current_node.get_left():
                        # print('move to LEFT, and dive to next level')
                        current_node = current_node.get_left()
                    else:
                        break

            if found_status:
                print("The data entry: {} found ".format(str(data)) + '#' * 30)
                # print('my parent node is '+ str(current_node.get_parent()))
            else:
                print("Attention! The data entry: {} is not found ".format(str(data)) + '#' * 30 + '\n')
            return found_status
        else:
            print("Attention! The data entry: {} is not found because the tree doesn't exist ".format(str(data))
                  + '#' * 30 + '\n' )
            return False

    def delete(self, data):
        """there are 3 possible scenarios:
        1. the node has no child
            delete the node and mark its parent node that 'node.next = None'
        2. the node has 1 child.
            delete the node and re-connect its parent node with its child node
        3. the node has 2 children
            find the Smallest key in the node's Right sub-tree
            replace the node with the Smallest key
        """
        current_node = self.root
        print('begin deleting data : {} '.format(data) + '#' * 50)
        if self.root:
            # Determine left/right side should be chosen for the new node
            found_status = False
            while not found_status:
                if data == current_node.get_data():
                    parent_node_data = current_node.get_parent().get_data()
                    print('Parent Node is ' + str(parent_node_data))
                    current_node = current_node.get_parent()
                    if data >= parent_node_data:
                        current_node.set_right(None)
                        print ('removing RIGHT')
                    else:
                        current_node.set_left(None)
                        print('removing LEFT')
                    found_status = True
                    break
                elif data > current_node.get_data():
                    if current_node.get_right():
                        # print('move to RIGHT, and dive to next level')
                        current_node = current_node.get_right()
                    else:
                        break  # no existing node larger than the current node.
                else:
                    if current_node.get_left():
                        # print('move to LEFT, and dive to next level')
                        current_node = current_node.get_left()
                    else:
                        break

            if found_status:
                print("The data entry: {} found and deleted ".format(str(data)) + '#' * 30)
                # print('my parent node is ' + str(current_node.get_parent()))
            else:
                print("Attention! The data entry: {} is not found ".format(str(data)) + '#' * 30 + '\n')
            return found_status
        else:
            print("Attention! The data entry: {} is not found because the tree doesn't exist ".format(str(data))
                  + '#' * 30 + '\n')
            return False

    def traverse_inOrder(self):
        """Steps:
        1 Go Left
        2 Process current node
        3 Go right
        """
        print('traversing tree(in-order)')
        tree_node = self.root
        result = []
        while not (tree_node == self.root and self.root.get_traversal_step() > 1) :
            if tree_node.get_traversal_step() < 3:
                print('\ncurrent node is {}'.format(tree_node.get_data()))
                print('steps: ' + str(tree_node.get_traversal_step()))
                print('Left child is: ' + str(tree_node.get_left()))  # for debugging
                # step1
                if tree_node.get_left():
                    tree_node.set_traversal_step()
                    while tree_node.get_left() and tree_node.get_left().get_traversal_step() < 3:
                        print('traversing to LEFT child')
                        tree_node = tree_node.get_left()
                        tree_node.set_traversal_step()
                else:
                      print('attempted to go LEFT but failed')
                      tree_node.set_traversal_step()

                # step2
                print('getting node data:' + str(tree_node.get_data()))
                result.append(tree_node.get_data())
                tree_node.set_traversal_step()

                #step3
                if tree_node.get_right():
                    print('traversing to RIGHT child')
                    tree_node.set_traversal_step()
                    tree_node = tree_node.get_right()
                else:
                    print('attempted to go RIGHT but failed')
                    tree_node.set_traversal_step()
            # step4 fall back to parent node
            else:
                if tree_node != self.root:
                    print('reversing to parent node {}'.format(tree_node.get_parent().get_data()))
                    tree_node = tree_node.get_parent()
        # step-final: reset all the step markers for the next traverse run.
        print(result)
        return result


    def traverse_preorder(self):
        level_result = []
        result = {}
        node = self.root
        if node:
            pass
        else:
            print('tree does not exist')
        return result

if __name__ == '__main__':
    INPUT_LIST = [50, 76, 21, 4, 32, 64, 15, 52, 14, 100, 83, 80, 2, 3, 70, 87]
    b = BinaryTree()
    for i in INPUT_LIST:
        b.insert(i)
    # print('Query match result : ' + str(b.query(87)))
    b.traverse_inOrder()
    b.query(3)
    b.delete(3)
    b.query(3)
    b.query(80)
    b.traverse_inOrder()
    b.traverse_inOrder()

1 个答案:

答案 0 :(得分:1)

我认为你做的事情比必要的要复杂得多。您可以使用递归函数的执行帧来跟踪哪些节点正在执行的操作:

def in_order_traversal(node):
    if node is None:
        return
    in_order_traversal(node.left)
    # do whatever you want to do on the current node here e.g.:
    print(node.data)
    in_order_traversal(node.right)

如果您不想使用递归,可以使用堆栈将相同的算法转换为迭代版本。这是一个使用list作为堆栈的版本,用于跟踪我们访问过的孩子的父节点,但是他们还没有自己处理:

def in_order_traversal_iterative(node):
    stack = []
    while node is not None or stack:
        while node is not None:
            stack.append(node)
            node = node.left
        node = stack.pop()
        print(node.data)  # process node
        node = node.right

这些实现都不需要修改节点,因此您可以根据需要多次运行它们,并且它们都能正常工作。

请注意,在我的示例代码中,我没有使用节点的get_Xset_Y方法。 Python中通常不需要访问器方法,公共属性更好。 getter和setter用于其他语言(如C ++和Java)的主要原因是允许您在不破坏类的公共API的情况下添加验证或更改属性的内部实现。在Python中,如果要添加验证或更改公共属性的实现,可以使用property将属性查找转换为方法调用。