Question

最终目标是将节点从一个树复制到另一个树。我想访问二叉树中的每个节点，并在经过多次遍历后返回一个节点实例。我似乎无法弄清楚如何返回特定节点。每次返回的节点都匹配根节点的id，因为我将根节点传递给函数。

class node():

    def __init__(self):
        self.parent = None
        self.left = None
        self.right = None

    def randnode(self, target):
        global count
        if target == count:
            # print 'At target', '.'*25, target
            return self
        else:
            if self.left is not None:
                count += 1
                self.left.randnode(target)
            if self.right is not None:
                count += 1
                self.right.randnode(target)

Answer 1

如果你正在进行DFS并计算迭代次数，你甚至不需要递归，只需要尝试一堆地方，然后弹出/推送数据。

def dfs(self,target):
    count = 0
    stack = [start]
    while stack:
        node = stack.pop()
        if count == target:
            return node

        if node is None:    # since we push left/right even if None
            continue        # stop pushing after we hit None node
        stack.extend([self.left,self.right])
    return -1  # ran out of nodes before count

加分点：将堆栈交换为BFS队列

除此之外，您可能希望将计数作为参数传递，就像所有自尊的递归调用一样，您可以使此无状态; - ）

class node():

    def __init__(self):
        self.parent = None
        self.left = None
        self.right = None

    def randnode(self, target,count=0):
        if target == count:
            # print 'At target', '.'*25, target
            return self
        if self.left is not None:
            return self.left.randnode(target,count + 1)
        if self.right is not None:
            return self.right.randnode(target,count + 1)

遍历树并在python中遍历n次后返回一个节点实例

1 个答案: