LeetCode 589 N-ary Tree Preorder Traversal

LeetCode

Given the root of an n-ary tree, return the preorder traversal of its nodes’ values.

Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]

Constraints:

The number of nodes in the tree is in the range [0, 10<sup>4</sup>]. 0 <= Node.val <= 10<sup>4</sup>
The height of the n-ary tree is less than or equal to 1000.

*Follow up: Recursive solution is trivial, could you do it iteratively?

题目大意：

求n个儿子的树的前序遍历

DFS解题思路：

严格按照定义，先root，再加入儿子节点

解题步骤：

N/A

注意事项：

1. root.children可能为None，所以要default成[]

Python代码：

def preorder(self, root: 'Node') -> List[int]:
	return self.dfs(root)

def dfs(self, root):
	if not root:
		return []
	res = [root.val]
	for child in root.children or []:
		res.extend(self.dfs(child))
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)

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迭代算法II解题思路：

用stack，但不能用模板，因为太多儿子，所以类似于Iterator方法Leetcode 341 ，先把该层的儿子节点反着加入到stack，保证后加入的后遍历。

Python代码：

def iterative_preorder(self, root: 'Node') -> List[int]:
	if not root:
		return []
	res, stack = [], []
	stack.append(root)

	while stack:
		node = stack.pop()
		res.append(node.val)
		for child in reversed(node.children or []):
			stack.append(child)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)。