Question

我需要以下代码的帮助来回答。我正在尝试使用堆栈而不是递归在n元树上执行后序遍历，因为python的限制为1000个递归。我在极客身上找到了相同“ https://www.geeksforgeeks.org/iterative-preorder-traversal-of-a-n-ary-tree/”的预遍历代码。但是我无法将其隐藏到后期订购中。任何帮助都会很棒。

Answer 1

这是我使用的iteration的{{1}}版本：

stack

测试代码和输出：

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.children = []

def postorder_traversal_iteratively(root: 'TreeNode'):
    if not root:
        return []
    stack = [root]
    # used to record whether one child has been visited
    last = None

    while stack:
        root = stack[-1]
        # if current node has no children, or one child has been visited, then process and pop it
        if not root.children or last and (last in root.children):
            '''
            add current node logic here
            '''
            print(root.val, ' ', end='')

            stack.pop()
            last = root
        # if not, push children in stack
        else:
            # push in reverse because of FILO, if you care about that
            for child in root.children[::-1]:
                stack.append(child)

视觉n元树和输出：

n1 = TreeNode(1)
n2 = TreeNode(2)
n3 = TreeNode(3)
n4 = TreeNode(4)
n5 = TreeNode(5)
n6 = TreeNode(6)
n7 = TreeNode(7)
n8 = TreeNode(8)
n9 = TreeNode(9)
n10 = TreeNode(10)
n11 = TreeNode(11)
n12 = TreeNode(12)
n13 = TreeNode(13)

n1.children = [n2, n3, n4]
n2.children = [n5, n6]
n4.children = [n7, n8, n9]
n5.children = [n10]
n6.children = [n11, n12, n13]

postorder_traversal_iteratively(n1)

进行后期订购的另一个想法是更改结果，例如将结果插入标题。

效率较低，但易于编码。您可以在here

中找到一个版本

我在github中为上述算法总结了1 / | \ / | \ 2 3 4 / \ / | \ 5 6 7 8 9 / / | \ 10 11 12 13 # output: 10 5 11 12 13 6 2 3 7 8 9 4 1。
如果您对以下内容感兴趣，请观看： https://github.com/recnac-itna/Code_Templates

一元树的后序遍历

1 个答案: