LeetCode 589. N-ary Tree Preorder Traversal

原题链接在这里：https://leetcode.com/problems/n-ary-tree-preorder-traversal/

题目：

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, 104].
• 0 <= Node.val <= 104
• 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?

题解：

Could use top down recursion.

Time Complexity: O(n). n is the number of nodes in the tree.

Space: O(logn). regardless res. stack space.

AC Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        dfs(root, res);
        return res;
    }
    
    private void dfs(Node root, List<Integer> res){
        if(root == null){
            return;
        }
        
        res.add(root.val);
        for(Node next : root.children){
            dfs(next, res);
        }
    }
}

AC C++:

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> res;
        dfs(root, res);
        return res;
    }

private: 
    void dfs(Node* root, vector<int>& res){
        if(!root){
            return;
        }

        res.push_back(root->val);
        for(Node* child : root->children){
            dfs(child, res);
        }
    }
};

AC Python:

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def preorder(self, root: 'Node') -> List[int]:
        res = []
        self.dfs(root, res)
        return res
    def dfs(self, root, res):
        if not root:
            return
        res.append(root.val)
        for child in root.children:
            self.dfs(child, res)

Could also do it iteratively. It is the method 2 in Binary Tree Preorder Traversal.

Time Complexity: O(n).

Space: O(logn). stack space.

AC Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        Stack<Node> stk = new Stack<>();
        stk.push(root);
        while(!stk.isEmpty()){
            Node cur = stk.pop();
            res.add(cur.val);
            for(int i = cur.children.size() - 1; i >= 0; i--){
                if(cur.children.get(i) != null){
                    stk.push(cur.children.get(i));
                }    
            }
        }
        
        return res;
    }
}

AC C++:

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> res;
        if(!root){
            return res;
        }

        stack<Node*> stk;
        stk.push(root);
        while(!stk.empty()){
            Node* top = stk.top();
            stk.pop();
            res.push_back(top->val);
            for(int i = top->children.size() - 1; i >= 0; i--){
                if(top->children[i]){
                    stk.push(top->children[i]);
                }
            }
        }

        return res;
    }
};

AC Python:

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def preorder(self, root: 'Node') -> List[int]:
        res = []
        if not root:
            return res
        stk = [root]
        while stk:
            cur = stk.pop()
            res.append(cur.val)
            stk.extend(child for child in cur.children[::-1] if child is not None)
        return res

类似N-ary Tree Postorder Traversal, Binary Tree Preorder Traversal.