N叉树的后序遍历

Question

我需要以下代码的帮助来回答。我试图使用堆栈而不是递归在n进制树上执行postorder遍历，因为python有1000个递归限制。我为极客在极客身上找到了相同的"https://www.geeksforgeeks.org/iterative-preorder-traversal-of-a-n-ary-tree/“命令遍历的代码。但我无法把它转到邮购处去。任何帮助都会很好。

Answer 1

下面是我使用的带有iteration版本的stack：

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)

测试代码和输出：

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)

可视化n叉树和输出：

                   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  

另一种方法是修改结果，例如将结果插入头。

它效率较低，但易于编码。您可以在这里中找到一个版本

我在我的github中总结了像上面这样的算法的code templates。

如果您对它感兴趣，请注意：模板