0589. N ary Tree Preorder Traversal - kumaeki/LeetCode GitHub Wiki

589. N-ary Tree Preorder Traversal

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?

解法1

N-ary Tree 和普通的Binary Tree一样,只是多了端点, 少了顺序.但是在前序输出的时候, 算法都是一样的. 题目中也提到了, 最简单的就是递归.

/*
// 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> list = new ArrayList<Integer>();
        helper(root, list);
        return list;
    }
    
    private void helper(Node node, List<Integer> list){
        if(node == null)
            return;
        
        list.add(node.val);
        for(Node n : node.children)
            helper(n , list);
    }
}

解法2

用队列来存储子节点

/*
// 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> list = new ArrayList<Integer>();
        if(root == null)
            return list;

        ArrayDeque<Node> que = new ArrayDeque<Node>();
        que.add(root);
        while(!que.isEmpty()){
            Node node = que.removeFirst();
            list.add(node.val);
            if(node.children != null && !node.children.isEmpty()){
                for(int i = node.children.size() - 1; i >= 0; i--)
                    que.addFirst(node.children.get(i));
            }
        }
        return list;
    }
}
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