python - 在 Python 中使用迭代而不是递归遍历二叉树

Question

在下面的二叉树中，只有叶子保持值，没有内部节点保持值。我使用递归实现了遍历(计算保留值的总和):

class Node:
    def new(rep):
        if type(rep) is list:
            left = Node.new(rep[0])
            right = Node.new(rep[1])
            return Internal(left, right)
        else:
            return Leaf(rep)


class Leaf(Node):
    def __init__(self, val):
        self.val = val

    def sum_leaves(self, sum):
        return sum + self.val


class Internal(Node):
    def __init__(self, left, right):
        self.left = left
        self.right = right

    def sum_leaves(self, sum):
        return self.right.sum_leaves(self.left.sum_leaves(sum))


class Tree:
    def __init__(self, rep):
        self.root = Node.new(rep)

    # Traverse the tree and return the sum of all leaves
    def sum_leaves(self):
        return self.root.sum_leaves(0)


treerep = [[3, [5, -1]], [[1, 7], [2, [3, [11, -9]]]]]
tree = Tree(treerep)
print(tree.sum_leaves())

这种情况下的输出是:

22

如何使用 iteration 而不是 sum_leaves 方法的递归？

Answer 1

您可以使用使用 while 循环的深度优先搜索:

class Tree:
    def __init__(self, rep):
        self.root = Node.new(rep)

    def sum_dfs(self):

        sum = 0
        stack = [self.root]

        while len(stack):

            node = stack.pop()

            if isinstance(node, Internal):
                stack.append(node.left)
                stack.append(node.right)

            elif isinstance(node, Leaf):
                sum += node.val

        return sum