Python实现二叉树的遍历

Python实现二叉树的遍历

class BinaryTree(object):
    def __init__(self, value=None, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

    def rebuild(self, preOrder, inOrder):
        """
        根据前序列表和中序列表,重建二叉树
        :param preOrder: 前序列表
        :param inOrder: 中序列表
        :return: 二叉树
        """
        if preOrder == None or inOrder == None or len(preOrder) <=0 or len(inOrder) <=0                 or len(preOrder) != len(inOrder):
            return None
        cur = BinaryTree(preOrder[0])
        index = inOrder.index(preOrder[0])
        cur.left = self.rebuild(preOrder[1: index+1], inOrder[:index])
        cur.right = self.rebuild(preOrder[index+1:], inOrder[index+1:])
        return cur

    def preOrder(self, tree):
        """
        前序遍历
        :param tree:
        :return:
        """
        if tree == None:
            return None
        print(tree.value, end=‘ ‘)
        self.preOrder(tree.left)
        self.preOrder(tree.right)

    def preOrderLoop(self, tree):
        """
        前序遍历的循环实现
        :param tree:
        :return:
        """
        if tree == None:
            return None
        lst = []
        node = tree
        while node != None or len(lst) > 0:
            if node != None:
                lst.append(node)
                print(node.value, end=‘ ‘)
                node = node.left
            else:
                item = lst[len(lst)-1]
                lst = lst[:-1]
                node = item.right

    def inOrder(self, tree):
        """
        中序遍历
        :param tree:
        :return:
        """
        if tree == None:
            return None
        self.inOrder(tree.left)
        print(tree.value, end=‘ ‘)
        self.inOrder(tree.right)

    def inOrderLoop(self, tree):
        """
        中序遍历循环实现
        :param tree:
        :return:
        """
        if tree == None:
            return
        lst = []
        node = tree

        while node != None or len(lst) > 0:
            if node != None:
                lst.append(node)
                node = node.left
            else:
                item = lst[len(lst)-1]
                lst = lst[:-1]
                print(item.value, end=‘ ‘)
                node = item.right

    def postOrder(self, tree):
        """
        后序遍历
        :param tree:
        :return:
        """
        if tree == None:
            return None
        self.postOrder(tree.left)
        self.postOrder(tree.right)
        print(tree.value, end=‘ ‘)

    def postOrderLoop(self, tree):
        """
        后续遍历的循环实现
        :param tree:
        :return:
        """
        if tree == None:
            return None

        visited = set()
        lst = []
        node = tree

        while node != None or len(lst) > 0:
            if node != None:
                lst.append(node)
                node = node.left
            else:
                item = lst[len(lst)-1]
                if item.right != None and item.right not in visited:
                    node = item.right
                else:
                    print(item.value, end=‘ ‘)
                    visited.add(item)
                    lst = lst[:-1]

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Python实现二叉树的遍历

原文：http://abelxu.blog.51cto.com/9909959/1968898