[英]Postorder traversal of an n ary tree
我需要以下代码的帮助才能回答。 我正在尝试使用堆栈而不是递归对 n 元树执行后序遍历,因为 python 有 1000 次递归的限制。 我在 geeks for geeks 上找到了相同的“ https://www.geeksforgeeks.org/iterative-preorder-traversal-of-an-ary-tree/ ”的预序遍历代码。 但我无法将其转换为后期订购。 任何帮助都会很棒。
这是我使用的带有stack
的iteration
版本:
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)
视觉 n 元树和输出:
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
另一个做后序的想法是改变结果,比如将结果插入到头部。
它效率较低,但易于编码。 你可以在这里找到一个版本
我在我的 github 中总结了上述算法的code templates
。
有兴趣的可以看看: https : //github.com/recnac-itna/Code_Templates
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